proving that $n^n \le (n!)^2$ I want to prove that $n^n \le (n!)^2$.
Now I tried by induction:
for $n=1$ ,$1=1$ and $P(1)$ is true
I suppose that $P(n)$ is true and I have to demonstrate that $P(n+1)$ is true
$$((n+1)!)^2=(n+1)^2*(n!)^2 \le (n+1)^2*n^n=(n+1)(n+1)*n^n=(n+1)^n$$
But I'm not sure about the last passage


Before editing for $n^n \ge (n!)^2$ 

Note that it is false indeed
$$n^n=\overbrace{n\cdot n \cdot n\cdot...\cdot n} ^{n \, terms} $$
$$(n!)^2=(\overbrace{n\cdot (n-1) \cdot (n-2)\cdot...\cdot 1}^{n \, terms})^2\ge(n\cdot (n-1) \cdot (n-2)\cdot...\cdot 1)\cdot n\ge n^n$$

After editing for $n^n \le (n!)^2$

induction step
$$(n+1)^{n+1}=n^n(n+1)\frac{(n+1)^n}{n^n}=n^n(n+1)\left(1+\frac1n\right)^n\stackrel{Ind. Hyp.}\le(n!)^2(n+1)e \le ((n+1)!)^2$$
which is true for $n\ge 2$.
