Proving that $n!≤((n+1)/2)^n$ by induction I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this: 
$V(1): 1≤1 \text{  true}$ 
$V(n): n!≤((n+1)/2)^n$ 
$V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$
and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$

It is more easy to prove this inequality without induction. Really $$0 < i\cdot (n + 1 - i) = \left(\frac{n+1}2 + \frac{2i - n - 1}2\right)\left(\frac{n+1}2 - \frac{2i - n - 1}2\right) = \left(\frac{n+1}2\right)^2 - \left(\frac{2i - n - 1}2\right)^2 \le \left(\frac{n+1}2\right)^2.$$
Multiply this inequalities for all $i = 1, 2, \ldots, \left\lfloor\frac n2\right\rfloor$ and by $\frac{n+1}2 = \frac{n+1}2$ for odd $n$ to get $n! \le \left(\frac{n+1}2\right)^n$ as desired.
