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

Notice:  $(n!) = (1*2*...... *n)^2 = (1*2*...... *n)*(1*2*...... *n)=$
$(1*n) * (2*(n-1)) * .......  * (n*1)$.
If we can prove that $k*(n-k+1) \ge n$ we will have
$(1*n) * (2*(n-1)) * .......  * (n*1) \ge n*n*n*....* = n^n$, and that will be it.
We can probably prove $k*(n-k+1) \ge n$ by induction but I'd rather note:
$k \le n$ and $k\ge 1$ so $k(n-k + 1) = kn - k^2 + k = n + (k-1)n -k^2 + k \ge n + (k-1)k - k^2 +k = n$.
.....
Intuitively this is also a result of AM-GM.  $k$ and $n-k + 1$ average to $\frac {n+1}2$ so if $m = \frac {n+1}2$ and $k = m - d$ and $n-k+1 = m + d$ we have $k(n-k + 1) = (m - d)(m+d) = m^2 - d^2 \ge m^2 = \frac {(n+1)^2}4$.  If $n\ge 3$ then $\frac {(n+1)^2}4 \ge \frac {(n+1)^2}{n+1} = n+ 1> n$.  And for $n=1,2,3$ then $\frac {(n+1)^2}4 = 1, \frac 94, 4 \ge n$.
