How can I show that $n! \leqslant \left(\frac{n+1}{2}\right)^n$? 
Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$

I know that it can be done by induction but I always find line where I do not know what to do next.

Hint:
$$
(n!)^2 = 1\times n \times 2\times (n-1) \times \dots
= \prod_{k=1}^n k(n+1-k)
$$
then use
$$
k\times (n+1 -k) =
\left(\frac {n+1}2 + \frac {n+1}2 - k\right)\left(\frac {n+1}2 - \frac {n+1}2 + k\right)
\\= \left(\frac {n+1}2\right)^2 - \left( \frac {n+1}2 - k\right)^2
\le \left(\frac {n+1}2\right)^2
$$
to get
$$
(n!)^2 \le \left(\frac {n+1}2\right)^{2n}
$$
