# Proving $(n+1)^n > 2^n n!$

If $n>1$ is a natural number, how to prove that

$$(n+1)^n > 2^n n!$$

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What have you tried? Your questions always read: "how to prove (fill in something homeworkish)." Please show a little bit of your own efforts and read meta.math.stackexchange.com/q/1803/5363 –  t.b. Aug 13 '12 at 4:55
Are you allowed to use Stirling? –  Guess who it is. Aug 13 '12 at 4:58
@J.M. This follows from AM GM –  Pedro Tamaroff Aug 13 '12 at 4:58
Stirling here seems like a pretty big hammer to squash a little bug. –  MJD Aug 13 '12 at 4:58
@Peter, I thought of that a few seconds after writing that comment. I'm getting lazy... –  Guess who it is. Aug 13 '12 at 5:02

By the AM GM inequality, \eqalign{ & \root n \of {n!} = \root n \of {1 \cdot 2 \cdot 3 \cdot 4 \cdots n} \leq \frac{{1 + 2 + 3 + \cdots + n}}{n} \cr & \frac{{\left( {n + 1} \right)n}}{{2n}} = \frac{{n + 1}}{2} \cr}

Then $$n! \leq {\left( {\frac{{n + 1}}{2}} \right)^n}$$ as desired.

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Prove it by induction on $n$. The base case $n=2$ is trivial. For the induction step you want to show that if $(n+1)^n>2^nn!$, then $(n+2)^{n+1}>2^{n+1}(n+1)!$. This will certainly be true if

$$\frac{(n+2)^{n+1}}{(n+1)^n}\ge \frac{2^{n+1}(n+1)!}{2^nn!}=2(n+1)\;.\tag{1}$$

The inequality $(1)$ is equivalent to the inequality

$$\left(\frac{n+2}{n+1}\right)^{n+1}=\left(1+\frac1{n+1}\right)^{n+1}\ge 2\;.$$

Now use the binomial theorem.

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@Peter: Yes, though I’m in the middle of writing up a couple of answers. –  Brian M. Scott Aug 13 '12 at 5:24

Let's consider the following auxiliary sequence: $$u_{n}=\frac{(n+1)^n}{2^n n!}$$ Then we notice the sequence is strictly increasing $$\frac{u_{n+1}}{u_{n}}=\frac{(n+2)^{n+1}}{2^{n+1} (n+1)!} \cdot \frac{2^n n!}{(n+1)^n}= \frac{1}{2} \left(1+\frac{1}{n+1}\right)^{n+1} > 1\space \ \tag1$$ and $$u_{2}= \frac{9}{8} \tag2$$

From $(1)$ and $(2)$ we get the needed inequality $$(n+1)^n > 2^n n!$$

Q.E.D.

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