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Prove that $n!>\left(\dfrac{n}{e}\right)^{n}$.

I used induction principle but cannot solve it for the $(m+1)$-th term after taking the $m$th term to be true.

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Some of the answers at… are related to this. – asmeurer Aug 25 '13 at 20:36

Here the key is to use the appropriate definition of $e^x$, namely:

$$e^x = \sum_{k=0}^{\infty}\frac{1}{k!}x^k$$

Plugging in $x = n$ we get

$$e^n = \sum_{k=0}^{\infty} \frac{1}{k!}n^k$$

and hence, breaking this sum up a little we get our inequality: $$n! e^n = n^n + \sum_{k\ne n} \frac{n!}{k!}n^k > n^n$$

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@PeterTamaroff There are many equivalent ways to express $e$, here by ``correct'' I mean most useful for the problem. I can edit if this is not clear – Deven Ware Aug 25 '13 at 17:58
Of course you mean that, I just think the use of the word "correct" is misleading. – Pedro Tamaroff Aug 25 '13 at 17:59
@PeterTamaroff Okay, edited. – Deven Ware Aug 25 '13 at 18:06

Hint: Show that $$ \ln(n!)=\sum_{k=1}^n \ln k >\int_1^n\ln x\, dx. $$

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That's a clever approach! Nice one! – dreamer Aug 25 '13 at 18:02

Inductively, if $n!>\frac{n^n}{e^n}$ and you multiply both sides by $n+1$, then you have that $(n+1)!>(n+1)\frac{n^n}{e^n}$, so it suffices to prove that $(n+1)\frac{n^n}{e^n}>\frac{(n+1)^{n+1}}{e^{n+1}}$. Can you continue from here?

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+1 This requires the least amount of machinery of the presented solutions, I think. – Jyrki Lahtonen Aug 25 '13 at 17:11

Hint: write out the series for $e^n$ and pick out a relevant term amongst the positive terms which make up the sum.

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You posted a hint and Deven independently posted a solution using the same idea => 10 vote difference. Isn't that always so? +1 to both of you from me has been there from the beginning, of course. – Jyrki Lahtonen Aug 25 '13 at 18:10

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