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Looking at the Wikipedia page on Mathematical Induction, I see that $n! < \frac{n^n}{2^n}\; \forall n>6$
I have been trying to prove that $n! < n^n \; \forall n>5$ using induction myself and my result is a clear special case of the above inequality. Does anyone have any hints on how to show either of these using induction? This came up when a student inquired about proving the convergence of a series, and induction seemed easier to show than a proof using Stirling's Approximation, but I can't seem to find a proof.

Brevan Ellefsen
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    Don't use induction, the result is clear if $n\ge 2$. For $n!$ is the product of $n$ numbers, all $\le n$, and some $\ne n$. And $n^n$ is the product of $n$ numbers all equal to $n$. – André Nicolas Dec 17 '15 at 23:09

5 Answers5

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Hint: Show $(n+1)n^n < (n+1)^{n+1}$.

Thomas Andrews
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Actually, $n!<n^n$ for $n>1$. The case $n=2$ is proved by inspection.

Suppose $n!<n^n$, for $n\ge2$; then $(n+1)!=(n+1)n!<(n+1)n^n$.

Moreover, from $n<n+1$, we deduce $n^n<(n+1)^n$. So we're done.

For $n!<n^n/2^n$, it's true only for $n>5$: indeed $6!=720$, while $6^6/2^6=729$.

So assume $n!<n^n/2^n$, for $n\ge6$. Then $$ (n+1)!=(n+1)n!<(n+1)\frac{n^n}{2^n} $$ and so we need to prove that $$ (n+1)\frac{n^n}{2^n}\le\frac{(n+1)^{n+1}}{2^{n+1}} $$ which is equivalent to $$ 2n^n\le (n+1)^n=n^n+\binom{n}{1}n^{n-1}+\sum_{k=2}^n\binom{n}{k}n^{n-k} $$ and this is clearly true.

egreg
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  • Beautiful second proof, thanks for helping! Not sure how I missed such a simple induction for the first proof, but the second is really interesting... nice use of the binomial theorem – Brevan Ellefsen Dec 18 '15 at 00:04
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Fill in the details:

$$(n+1)!=(n+1)n!<(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}$$

YoTengoUnLCD
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  • Thanks for your answer, I really thought it was the easiest to understand. Quick follow up, does the following hold as well as an extension of your proof?

    $$\text{Suppose } n! < n^{n-2}$$ $$(n+1)! = (n+1)n! < (n+1)n^{n-2} < (n+1)(n+1)^{n-2} = (n+1)^{n-1} $$

     – Brevan Ellefsen Dec 17 '15 at 23:56
  • Yes, it does! But only if you find the appropiate base case (for that, $n\geq 5$ suffices).

    I think it's even true that $n!<n^{n-m}$ for whatever fixed $m\in \Bbb N$ if you carefully pick your base case.

     – YoTengoUnLCD Dec 18 '15 at 00:16
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$$\underbrace{n!=1*2*3\; ...*\;n}_{n\text{ digits}}$$

$$\underbrace{n^n=n*n*n\;...*\;n}_{n\text{ digits}}$$

Start with the first digit in each and proceed until reaching $n$.

$$1<n$$ $$2<n$$ $$3<n$$ $$\vdots$$ $$n=n$$

This should make it obvious that $n!<n^n$.

Leonidas Lanier
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$(1+1+...+1)^{n}$ use multinomial expansion formula Then $(1+1+...+1)^{n}>{n\choose{1}{1}{...}{1}}$

  • haha, yeah. That would do it, even if it is sorta nuking the fly XD This post was a while back... can't say I need the help anymore, and I could definitely answer this myself now, but it's always fun to see a creative answer on an old post! – Brevan Ellefsen Aug 15 '17 at 04:50
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    I am the new user, seeing the post for the first time. Tried to give different answer. Then Binomiamial expansion clicked on my mind. Then I generalize it. –  Aug 15 '17 at 04:57