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My textbook says that: $$ \frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n<e^n $$ But I do not understand this. Can you please enlighten me?

Edit: How do you show that $$ \frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n $$

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4 Answers

For the left-hand equality: You can prove this by induction: The ratio of ${(n+1)^n \over n!}$ to ${n^{n-1} \over (n-1)!}$ is exactly $${(n-1)! \over n!} {(n+1)^n \over n^{n-1}}$$ $$= {1 \over n}{(n+1)^n \over n^{n-1}}$$ $$= {(n+1)^n \over n^n}$$ $$= (1 + {1 \over n})^n$$ This is exactly the $n$-th factor of the right-hand side.

For the right-hand inequality: You can do this using the inequality $\ln(1 + x) < x$ for $x > 0$, which can be proven for example by observing that the two functions $\ln(1 + x)$ and $x$ are equal when $x = 0$ while the derivative of $\ln(1 + x)$ is less than the derivative of $x$ (namely $1$) when $x \geq 0$.

Plugging in $x = 1/k$ for any $k$ you get $\ln(1 + {1 \over k}) < {1 \over k}$. Taking $e$ to both sides you get $1 + {1 \over k} < e^{1 \over k}$ which implies $(1 + {1 \over k})^k < e$. Multiplying all of these together from $k = 1$ to $k = n$ gives the inequality you seek.

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But how do you show that $\frac{(n+1)^n}{n!}=\cdots$? –  Vafa Khalighi Mar 27 '11 at 1:03
    
edited to include a proof of that too. –  Zarrax Mar 27 '11 at 1:23
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For the first part

note that each term is $$\begin{align} \left(\frac{1+k}{k}\right)^k =& \frac{(1+k)^k}{k^k}\\ =&\frac{1}{k}\frac{(1+k)^k}{k^{k-1}}\\ =&\frac{1}{k}\frac{(1+k)^k}{(1+(k-1))^{k-1}} \end{align}$$

so every denominator cancels nicely with the previous numerator:

$$ \begin{align}\frac{1}{n!}\left(\frac{(1+1)^1}{(1+0)^0}\frac{(1+2)^2}{(1+1)^1}\frac{(1+3)^3}{(1+2)^2}\cdots\frac{(1+n)^n}{n^{n-1}}\right)\\ =\frac{1}{n!}(1+n)^n\end{align}$$

For the second part

note that $f(k) = \left(1+\frac{1}{k}\right)^k < e$ for every $k>0$ since $f(k)$ is a monotonically increasing function and $\lim_{k\to\infty}f(k) = e$, so that the product of any $n$ of them is less than $ee\cdots e = e^n$.

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Thanks, how $k^{k-1}=(1+(k-1))^{k-1}$? –  Vafa Khalighi Mar 27 '11 at 2:00
    
This is so obvious. I do not why, I could not see it. Thanks –  Vafa Khalighi Mar 27 '11 at 2:05
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You can prove the equality by induction over $n\ge0$. The inequality stems from the fact that, for every $n\ge1$, $$ \left(1+\frac1n\right)^n<\mathrm{e}, $$ the simplest proof of which uses the fact that $1+x<\mathrm{e}^x$ for every positive real number $x$.

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use induction. you may easily show it is true for $n=1,2,3$ (though u just need for $n=1$). assume it is true for $n=m$ and with the help of this assumption show it is true for $n=m+1$

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