Show that $\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n$ I'd like to prove that  $\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n$. 
What I have so far:
$$\sqrt[n]{n!} = \sqrt[n]{1\cdot 2 \cdot \ldots \cdot n} \leq \frac{1+\ldots +n}{n}=\frac{(n+1)n}{2n}=\frac{(n+1)}{2}.$$
Thus
$$\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n.$$
But how do I go from $n+1$ to $n$?
 A: In the first line you've shown that
$$
n! \leq \left(\frac{n+1}{2}\right)^n,
$$
and the expression on the right is
$$
\left(\frac{n+1}{2}\right)^n = \left(\frac{n}{2}\right)^n \left(1+\frac{1}{n}\right)^n < \left(\frac{n}{2}\right)^n e.
$$
A: As obtained in the OP:
$$
\sqrt[n]{n!}\le \frac{n+1}{2},
$$
and hence
$$
n!\le \left(\frac{n+1}{2}\right)^{\!n}=2^{-n}n^n\left(\frac{n+1}{n}\right)^{\!n}=
\left(1+\frac{1}{n}\right)^{\!n}\left(\frac{n}{2}\right)^{\!n}
<\mathrm{e}\left(\frac{n}{2}\right)^{\!n},
$$
since
$$
\left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e},
$$
which is is due to the fact that
$$
1+\frac{1}{n}<\mathrm{e}^{1/n}.
$$
A: $n!<e\bigg(\dfrac{n}{2}\bigg)^n \implies \sqrt{2\pi n}\bigg(\dfrac{n}{e}\bigg)^n<e\bigg(\dfrac{n}{2}\bigg)^n\implies \dfrac{\sqrt{2\pi n}}{e}<\bigg(\dfrac{e}{2}\bigg)^n$  
The second step is by Stirling's Approximation. I hope you can prove the last inequality, which is according to me quite trivial to prove. If you are stuck try to use induction or try to use calculus.
