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

Question

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.