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

This is false for $n \ge 3$.
$27=3^3 < (3!)^2=36$.
Suppose $n^n <(n!)^2$
and $n \ge 3$.
Then
$\dfrac{(n+1)^{n+1}/((n+1)!^2)}{n^n/(n!)^2}
=(n+1)(1+1/n)^n/(n+1)^2
\lt e/(n+1)
\lt 1
$.
This shows that
$n^n/(n!)^2$
is a decreasing sequence.
