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?

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.
