TITLE: How to prove $n!\leq(\frac{n+1}{2})^n$
QUESTION [5 upvotes]: Prove that for $n\in\mathbb{N}$ $$n!\leq(\frac{n+1}{2})^n.$$
I've solved base case for $n=1$
$$1\leq(\frac{1+1}{2})^1=1$$
The second step I've made was that I assumed that $n!\leq(\frac{n+1}{2})^n$ And then I have $(\frac{n+2}{2})^{n+1}$. What do I need to do now?

REPLY [3 votes]: We start from
$$(n+1)!$$
the left side for n+1-th step. 
$$(n+1)!=(n+1) n!\leq (n+1) \frac{(n+1)^n}{2^n}$$
from n-th step. Rewrite:
$$(n+1) \frac{(n+1)^n}{2^n}=2(n+1) \frac{(n+1)^n}{2^{n+1}}$$
Noting that
$$2(n+1)^{n+1}\leq (n+2)^{n+1}$$
beause
$$\left(1+\frac{1}{n+1}\right)^{n+1}\geq 2$$
($(1+1/n)^n$ is an increasing sequence)
we have that
$$2(n+1) \frac{(n+1)^n}{2^{n+1}}\leq \left(\frac{n+2}{2}\right)^{n+1}$$