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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\end_preamble
-\use_default_options true
-\maintain_unincluded_children no
-\language english
-\language_package default
-\inputencoding utf8
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-\use_package amsmath 1
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-\cite_engine basic
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-\suppress_date false
-\justification true
-\use_refstyle 1
-\use_formatted_ref 0
-\use_minted 0
-\use_lineno 0
-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
-\paragraph_indentation default
-\is_math_indent 0
-\math_numbering_side default
-\quotes_style english
-\dynamic_quotes 0
-\papercolumns 1
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[00]
-\end_layout
-
-\end_inset
-
-How many ways are there to shuffle a 52-card deck?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $52!=80658175170943878571660636856403766975289505440883277824000000000000$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc2[10]
-\end_layout
-
-\end_inset
-
-In the notation of Eq.
- (2),
- show that 
-\begin_inset Formula $p_{n(n-1)}=p_{nn}$
-\end_inset
-
-,
- and explain why this happens.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $p_{n(n-1)}=n(n-1)\cdots(n-(n-1)+1)=n(n-1)\cdots2=n(n-1)\cdots1=p_{nn}$
-\end_inset
-
-.
- This is because 
-\begin_inset Formula $p_{n(n-1)}$
-\end_inset
-
- is the number of ways to take and arrange 
-\begin_inset Formula $n-1$
-\end_inset
-
- objects out 
-\begin_inset Formula $n$
-\end_inset
-
- and 
-\begin_inset Formula $p_{nn}$
-\end_inset
-
- is the number of ways to take and arrange the 
-\begin_inset Formula $n$
-\end_inset
-
- objects,
- which consists of first arranging 
-\begin_inset Formula $n-1$
-\end_inset
-
- objects out of the 
-\begin_inset Formula $n$
-\end_inset
-
- objects and then adding the one remaining.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc3[10]
-\end_layout
-
-\end_inset
-
-What permutations of 
-\begin_inset Formula $\{1,2,3,4,5\}$
-\end_inset
-
- would be constructed from the permutation 
-\begin_inset Formula $3\,1\,2\,4$
-\end_inset
-
- using Methods 1 and 2,
- respectively?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-
-\series bold
-Method 1:
-
-\series default
- 
-\begin_inset Formula $5\,3\,1\,2\,4$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,5\,1\,2\,4$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,1\,5\,2\,4$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,1\,2\,5\,4$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,1\,2\,4\,5$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-
-\series bold
-Method 2:
-
-\series default
- 
-\begin_inset Formula $4\,2\,3\,5\,1$
-\end_inset
-
-,
- 
-\begin_inset Formula $4\,1\,3\,5\,2$
-\end_inset
-
-,
- 
-\begin_inset Formula $4\,1\,2\,5\,3$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,1\,2\,5\,4$
-\end_inset
-
-,
- 
-\begin_inset Formula $3\,1\,2\,4\,5$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[13]
-\end_layout
-
-\end_inset
-
-Given the fact that 
-\begin_inset Formula $\log_{10}1000!=2567.60464...$
-\end_inset
-
-,
- determine exactly how many decimal digits are present in the number 
-\begin_inset Formula $1000!$
-\end_inset
-
-.
- What is the 
-\emph on
-most significant
-\emph default
- digit?
- What is the 
-\emph on
-least significant
-\emph default
- digit?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Given a number 
-\begin_inset Formula $x$
-\end_inset
-
- with 
-\begin_inset Formula $n+1$
-\end_inset
-
- digits 
-\begin_inset Formula $x_{n}x_{n-1}\cdots x_{1}x_{0}$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $x_{n}10^{n}\leq x<(x_{n}+1)10^{n}$
-\end_inset
-
- and therefore 
-\begin_inset Formula $n+\log_{10}x_{n}\leq\log x<n+\log_{10}(x_{n}+1)$
-\end_inset
-
-,
- with 
-\begin_inset Formula $0\leq\log_{10}X_{n}<\log_{10}(x_{n}+1)\leq1$
-\end_inset
-
-.
- Thus,
- 
-\begin_inset Formula $1000!$
-\end_inset
-
- has 2568 digits and the most significant digit is 4,
- since 
-\begin_inset Formula $\log_{10}4=0.60205...$
-\end_inset
-
- and 
-\begin_inset Formula $\log_{10}5=0.69897...$
-\end_inset
-
-.
- Furthermore,
- since 
-\begin_inset Formula $10\mid1000!$
-\end_inset
-
-,
- the least significant digit is 0.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc6[17]
-\end_layout
-
-\end_inset
-
-Using Eq.
- (8),
- write 
-\begin_inset Formula $20!$
-\end_inset
-
- as a product of prime factors.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Numbers up to 20 have prime factors up to 20,
- so the prime factors are 2,
- 3,
- 5,
- 7,
- 11,
- 13,
- 17,
- and 19.
- For the multiplicities,
-\begin_inset Formula 
-\begin{align*}
-\mu_{2} & =\sum_{k\geq1}\left\lfloor \frac{20}{2^{k}}\right\rfloor =10+5+2+1=18, & \mu_{11} & =\sum_{k\geq1}\left\lfloor \frac{20}{11^{k}}\right\rfloor =1,\\
-\mu_{3} & =\sum_{k\geq1}\left\lfloor \frac{20}{3^{k}}\right\rfloor =6+2=8, & \mu_{13} & =\sum_{k\geq1}\left\lfloor \frac{20}{13^{k}}\right\rfloor =1,\\
-\mu_{5} & =\sum_{k\geq1}\left\lfloor \frac{20}{5^{k}}\right\rfloor =4, & \mu_{17} & =\sum_{k\geq1}\left\lfloor \frac{20}{17^{k}}\right\rfloor =1,\\
-\mu_{7} & =\sum_{k\geq1}\left\lfloor \frac{20}{7^{k}}\right\rfloor =2, & \mu_{19} & =\sum_{k\geq1}\left\lfloor \frac{20}{19^{k}}\right\rfloor =1,
-\end{align*}
-
-\end_inset
-
-so 
-\begin_inset Formula $20!=2^{18}3^{8}5^{4}7^{2}\cdot11\cdot13\cdot17\cdot19$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc7[M10]
-\end_layout
-
-\end_inset
-
-Show that the 
-\begin_inset Quotes eld
-\end_inset
-
-generalized termial
-\begin_inset Quotes erd
-\end_inset
-
- function in Eq.
- (10) satisfies the identity 
-\begin_inset Formula $x?=x+(x-1)?$
-\end_inset
-
- for all real numbers 
-\begin_inset Formula $x$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $x+(x-1)?=x+\frac{1}{2}(x-1)x=\frac{1}{2}2x+\frac{1}{2}(x-1)x=\frac{1}{2}(x+1)x=x?$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc9[M10]
-\end_layout
-
-\end_inset
-
-Determine the values of 
-\begin_inset Formula $\Gamma(\frac{1}{2})$
-\end_inset
-
- and 
-\begin_inset Formula $\Gamma(-\frac{1}{2})$
-\end_inset
-
-,
- given that 
-\begin_inset Formula $(\frac{1}{2})!=\sqrt{\pi}/2$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{align*}
-(\tfrac{1}{2})!=\tfrac{1}{2}\Gamma(\tfrac{1}{2}) & \implies\Gamma(\tfrac{1}{2})=2(\tfrac{1}{2})!=\sqrt{\pi},\\
--\tfrac{1}{2}\Gamma(-\tfrac{1}{2})=\Gamma(\tfrac{1}{2}) & \implies\Gamma(-\tfrac{1}{2})=-2\Gamma(\tfrac{1}{2})=-2\sqrt{\pi}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[HM20]
-\end_layout
-
-\end_inset
-
-Does the identity 
-\begin_inset Formula $\Gamma(x+1)=x\Gamma(x)$
-\end_inset
-
- hold for all real numbers 
-\begin_inset Formula $x$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-No,
- as it doesn't hold when 
-\begin_inset Formula $\Gamma(x)$
-\end_inset
-
- or 
-\begin_inset Formula $\Gamma(x+1)$
-\end_inset
-
- are not defined.
- However,
- when they are defined,
-\begin_inset Formula 
-\begin{align*}
-x\Gamma(x) & =x\lim_{m}\frac{m^{x}m!}{\prod_{i=0}^{m}(x+i)}=\lim_{m}\frac{m^{x}m!}{\prod_{i=1}^{m}(x+i)}=\lim_{m}\frac{(m+1)^{x}(m+1)!}{\prod_{i=1}^{m+1}(x+i)}=\\
- & =\lim_{m}\frac{(m+1)^{x+1}m!}{\prod_{i=0}^{m}(x+1+i)}=\lim_{m}\left(\frac{m^{x+1}m!}{\prod_{i=0}^{m}((x+1)+i)}\left(\frac{m+1}{m}\right)^{x+1}\right)=\Gamma(x+1)\cdot1.
-\end{align*}
-
-\end_inset
-
-Note that,
- if 
-\begin_inset Formula $x\in\mathbb{Z}^{-}$
-\end_inset
-
-,
- then the terms from 
-\begin_inset Formula $m=-x$
-\end_inset
-
- onward are not defined,
- so this is not the scenario we are dealing with.
- For ant other value,
- this holds.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc11[M15]
-\end_layout
-
-\end_inset
-
-Let the representation of 
-\begin_inset Formula $n$
-\end_inset
-
- in the binary system be 
-\begin_inset Formula $n=2^{e_{1}}+2^{e_{2}}+\dots+2^{e_{r}}$
-\end_inset
-
-,
- where 
-\begin_inset Formula $e_{1}>e_{2}>\dots>e_{r}\geq0$
-\end_inset
-
-.
- Show that 
-\begin_inset Formula $n!$
-\end_inset
-
- is divisible by 
-\begin_inset Formula $2^{n-r}$
-\end_inset
-
- but not by 
-\begin_inset Formula $2^{n-r+1}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-This is exactly to say that,
- for 
-\begin_inset Formula $p=2$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mu=n-r$
-\end_inset
-
-.
- But 
-\begin_inset Formula $r$
-\end_inset
-
- is the sum of all bits of 
-\begin_inset Formula $n$
-\end_inset
-
- in base 2,
- so by Exercise 12,
- 
-\begin_inset Formula $\mu=\frac{1}{2-1}(n-r)=n-r$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc12[M22]
-\end_layout
-
-\end_inset
-
-(A.
- Legendre,
- 1808.) Generalizing the result of the previous exercise,
- let 
-\begin_inset Formula $p$
-\end_inset
-
- be a prime number,
- and let the representation of 
-\begin_inset Formula $n$
-\end_inset
-
- in the 
-\begin_inset Formula $p$
-\end_inset
-
--ary number system be 
-\begin_inset Formula $n=a_{k}p^{k}+a_{k-1}p^{k-1}+\dots+a_{1}p+a_{0}$
-\end_inset
-
-.
- Express the number 
-\begin_inset Formula $\mu$
-\end_inset
-
- of Eq.
- (8) in a simple formula involving 
-\begin_inset Formula $n$
-\end_inset
-
-,
- 
-\begin_inset Formula $p$
-\end_inset
-
-,
- and 
-\begin_inset Formula $a$
-\end_inset
-
-'s.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The question makes sense for 
-\begin_inset Formula $n\in\mathbb{N}^{*}$
-\end_inset
-
-.
- Then,
- given that the previous exercise showed that,
- for 
-\begin_inset Formula $p=2$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mu=n-r$
-\end_inset
-
-,
- where 
-\begin_inset Formula $r\coloneqq|\{k\in\mathbb{N}\mid a_{k}\neq0\}|=\sum_{k}a_{k}$
-\end_inset
-
-,
- a guess is that 
-\begin_inset Formula $\mu=n-\sum_{k}a_{k}$
-\end_inset
-
- for any other positive prime integer 
-\begin_inset Formula $p$
-\end_inset
-
- as well,
- that is,
- 
-\begin_inset Formula $n-\mu=\sum_{k}a_{k}$
-\end_inset
-
-.
- To prove this,
-\begin_inset Formula 
-\[
-\mu=\sum_{j\geq1}\left\lfloor \frac{n}{p^{j}}\right\rfloor =\sum_{j=1}^{k}\sum_{i=j}^{k}a_{i}p^{i-j}=\sum_{1\leq j\leq i\leq k}a_{i}p^{i-j}=\sum_{i=1}^{k}a_{i}\left(\sum_{j=0}^{i-1}p^{j}\right)=\sum_{i=1}^{k}\frac{a_{i}(p^{i}-1)}{p-1},
-\]
-
-\end_inset
-
-where the latter identity is because of the cyclotomic formula.
- Then,
-\begin_inset Formula 
-\begin{align*}
-\sum_{i=1}^{k}\frac{a_{i}(p^{i}-1)}{p-1} & =\frac{1}{p-1}\left(\sum_{i=1}^{k}a_{i}p^{i}-\sum_{i=1}^{k}a_{i}\right)=\frac{1}{p-1}\left((n-a_{0})-\sum_{i\geq1}a_{i}\right)\\
- & =\frac{1}{p-1}\left(n-\sum_{i\geq0}a_{i}\right).
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc22[HM20]
-\end_layout
-
-\end_inset
-
-Try to put yourself in Euler's place,
- looking for a way to generalize 
-\begin_inset Formula $n!$
-\end_inset
-
- to noninteger values of 
-\begin_inset Formula $n$
-\end_inset
-
-.
- Since 
-\begin_inset Formula $(n+\frac{1}{2})!/n!$
-\end_inset
-
- times 
-\begin_inset Formula $((n+\frac{1}{2})+\frac{1}{2})!/(n+\frac{1}{2})!$
-\end_inset
-
- equals 
-\begin_inset Formula $(n+1)!/n!=n+1$
-\end_inset
-
-,
- it seems natural that 
-\begin_inset Formula $(n+\frac{1}{2})!/n!$
-\end_inset
-
- should be approximately 
-\begin_inset Formula $\sqrt{n}$
-\end_inset
-
-.
- Similarly,
- 
-\begin_inset Formula $(n+\frac{1}{3})!/n!$
-\end_inset
-
- should be 
-\begin_inset Formula $\approx\sqrt[3]{n}$
-\end_inset
-
-.
- Invent a hypothesis about the ratio 
-\begin_inset Formula $(n+x)!/n!$
-\end_inset
-
- as 
-\begin_inset Formula $n$
-\end_inset
-
- approaches infinity.
- Is your hypothesis correct when 
-\begin_inset Formula $x$
-\end_inset
-
- is an integer?
- Does it tell anything about the appropriate value of 
-\begin_inset Formula $x!$
-\end_inset
-
- when 
-\begin_inset Formula $x$
-\end_inset
-
- is not an integer?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The previous examples show 
-\begin_inset Formula $\frac{(n+\frac{1}{m})!}{n!}\approx\sqrt[m]{n}=n^{\frac{1}{m}}$
-\end_inset
-
-,
- which is to say that 
-\begin_inset Formula $\frac{(n+x)!}{n!}\approx n^{x}$
-\end_inset
-
- for 
-\begin_inset Formula $x=\frac{1}{m}$
-\end_inset
-
-.
- As 
-\begin_inset Formula $n$
-\end_inset
-
- gets bigger,
- the approximation should probably be more accurate for the same value of 
-\begin_inset Formula $x$
-\end_inset
-
-,
- so we could see that 
-\begin_inset Formula $(n+x)!\approx n^{x}n!$
-\end_inset
-
-.
- For a non-negative integer 
-\begin_inset Formula $x$
-\end_inset
-
-,
- 
-\begin_inset Formula $(n+x)!=(n+1)\cdots(n+x)\cdot n!$
-\end_inset
-
-,
- so as 
-\begin_inset Formula $n$
-\end_inset
-
- gets bigger the approximation is more and more precise,
- and in fact,
-\begin_inset Formula 
-\[
-\lim_{n}\frac{n^{x}n!}{(n+x)!}=\lim_{n}\frac{n^{x}}{(n+1)\cdots(n+x)}=\lim_{n}\prod_{k=1}^{x}\frac{n}{n+k}=1.
-\]
-
-\end_inset
-
-Multiplying by 
-\begin_inset Formula $x!$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-x!=\lim_{n}\frac{n^{x}x!n!}{(n+x)!}=\lim_{n}(n^{x}n!)\frac{x!}{(n+x)!}=\lim_{n}\frac{n^{x}n!}{(x+1)\cdots(x+n)},
-\]
-
-\end_inset
-
-which is Euler's factorial formula.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc24[HM21]
-\end_layout
-
-\end_inset
-
-Prove the handy inequalities
-\begin_inset Formula 
-\begin{align*}
-\frac{n^{n}}{\text{e}^{n-1}} & \leq n!\leq\frac{n^{n+1}}{\text{e}^{n-1}}, & \text{integer }n & \geq1.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(Looking at the answers.) 
-\end_layout
-
-\end_inset
-
-We have
-\begin_inset Formula 
-\begin{align*}
-\frac{n^{n}}{n!} & =\prod_{k=1}^{n}\frac{n}{k}=\prod_{k=1}^{n-1}\frac{n}{k}=\prod_{k=1}^{n-1}\prod_{j=k}^{n-1}\frac{j+1}{j}=\prod_{1\leq k\leq j<n}\frac{j+1}{j}=\prod_{j=1}^{n-1}\left(\frac{j+1}{j}\right)^{j}=\\
- & =\prod_{j=1}^{n-1}\left(1+\frac{1}{j}\right)^{j}\leq\prod_{j=1}^{n-1}\left(\text{e}^{1/j}\right)^{j}=\text{e}^{n-1},
-\end{align*}
-
-\end_inset
-
-where for the inequality we use that 
-\begin_inset Formula $1+x\leq\text{e}^{x}$
-\end_inset
-
- for all real 
-\begin_inset Formula $x$
-\end_inset
-
-.
- Likewise,
-\begin_inset Formula 
-\begin{align*}
-\frac{n^{n+1}}{n!} & =n\frac{n^{n}}{n!}=n\prod_{j=1}^{n-1}\left(\frac{j+1}{j}\right)^{j}=\prod_{j=1}^{n-1}\frac{j+1}{j}\prod_{j=1}^{n-1}\left(\frac{j+1}{j}\right)^{j}=\prod_{j=1}^{n-1}\left(\frac{j+1}{j}\right)^{j+1}=\\
- & =\prod_{j=2}^{n}\left(\frac{j}{j-1}\right)^{j}=\prod_{j=2}^{n}\left(1-\frac{1}{j}\right)^{-j}\geq\prod_{j=2}^{n}\left(\text{e}^{-1/j}\right)^{-j}=\text{e}^{n-1}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document