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 : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$

An induction proof:
First, let's make it a little bit more eye-candy:
$$
n! \cdot 2^{n} \leq (n+1)^n
$$
Now, for $n=1$ the inequality holds. For $n=k\in\mathbb{N}$ we know that:
$$
k! \cdot 2^{k} \leq (k+1)^k
$$
holds and we need to prove:
$$
(k+1)! \cdot 2^{k+1} \leq (k+2)^{k+1}
$$
We will now prove this chain of inequalities (which gives us the actual proof):
$$
(k+1)! \cdot 2^{k+1} \leq 2(k+1)^{k+1} \leq (k+2)^{k+1}
$$
The first inequality is from the assumption (both sides multiplied by $2(k+1)$). Now we just need to prove the second one. In other words, we need to prove this (for some big enough positive integer $p$):
$$
2p^{p} \leq (p+1)^{p}
$$
And that's rather obvious. The inequality
$$
2 \leq \left(1+\frac{1}{p}\right)^{p}
$$
holds because the function on the right is known to be increasing and its limit (as $p\to\infty$) is $e$. So at some point on it has to be greater than 2.
