1. For n=1 and 2 it's equal and for n=3 left side is greater than right side. Let's assume for n=k it is also true.
$$(k+1)!^2=k!^2 \cdot (k+1)^2 \ge k^k \cdot (k+1)^2$$
Now, prove $k^k \ge (k+1)^{k-1}$. It is easy. :spin:

2. I don't think it is only for $n \in \mathbb{N}$. If it was then proving $n^n+A>n^2+n$ would be too easy.