How to prove $n!\leq(\frac{n+1}{2})^n$ 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?
 A: Just write that
$$
(n!)^2 = \prod_{k=1}^n k(n+1-k) \le \prod_{k=1}^n \left[\frac 12 (k+(n+1-k))\right]^2
= \left(\frac{n+1}2\right)^{2n}
$$
A: The induction proof can be made to work; use the fact that
$$\begin{align*}
\frac{\left(\frac{n+2}2\right)^{n+1}}{\left(\frac{n+1}2\right)^n}&=\frac{n+2}2\left(\frac{n+2}{n+1}\right)^n\\
&=\frac{n+2}2\left(1+\frac1{n+1}\right)^n\\\\
&>\frac{n+2}2\left(1+\frac{n}{n+1}\right)\\\\
&=\frac{(n+2)(2n+1)}{2n+2}\\\\
&=\frac{2n^2+5n+2}{2n+2}\\\\
&=n+\frac{3n+2}{2n+2}\\\\
&>n+1\;.
\end{align*}$$
The first inequality follows from the binomial theorem, for instance.
A: 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
$$2(n+1) \frac{(n+1)^n}{2^{n+1}}\leq \left(\frac{n+2}{2}\right)^{n+1}$$
A: Applying the AM-GM inequality you obtain $$\sqrt[n]n!=\sqrt[n]{1\cdot2\cdot\ldots\cdot(n-1)\cdot n}\overset{\text{AM-GM}}\le\frac{1+2+\ldots+(n-1)+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$$ Now raise both sides to the $n$-power to conclude.
