Comparing $n^n$ and $n!^2$ I tried to prove that if $n>2$ then $(n!)^2>n^n$ but did not managed. That is the trick to compare those as both grows rapidly? Induction seems hard: $((n+1)!)^2=(n+1)^2(n!)^2>(n+1)^2n^n$ but why $(n+1)^2n^n>(n+1)^{n+1}$? I also noted that $n^n=e^{n\ln n}$ and $n!^2=e^{2\ln n!}=e^{2\sum_{i=1}^n \ln i}$ but got stuck.

Suppose
$(n!)^2 > n^n$.
Then
$((n+1)!)^2
=(n!)^2(n+1)^2
>n^n(n+1)^2
$
so we need
$n^n(n+1)^2
\ge (n+1)^{n+1}
$
which is the same as
$n^n
\ge (n+1)^{n-1}
$
or,
dividing by
$n^{n-1}$,
$n 
\ge (1+1/n)^{n-1}
$.
Multiplying by
$1+1/n$,
this becomes
$n+1
\ge (1+1/n)^n
$.
This is true because
$(1+1/n)^n
< e
$,
as has been shown many times here.
Here is one of the easier:
Using the binomial theorem,
$\begin{array}\\
(1+1/n)^n
&=\sum_{k=0}^n \binom{n}{k}(1/n)^k\\
&=\sum_{k=0}^n \frac{n!}{k!(n-k)!}(1/n)^k\\
&=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(n-j)}{k!n^k}\\
&=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(1-j/k)}{k!}\\
&<\sum_{k=0}^n \frac{1}{k!}\\
&<\sum_{k=0}^{\infty} \frac{1}{k!}\\
&= e
\end{array}
$
Actually,
all we need is
a bound on 
$\sum_{k=0}^{\infty} \frac{1}{k!}
$.
An easy one
is gotten from
$k! \ge 2\cdot 2^{k-2}$
for $k \ge 2$,
easily proved by induction.
Then
$\begin{array}\\
\sum_{k=0}^{\infty} \frac{1}{k!}
&=1+1+\sum_{k=2}^{\infty} \frac{1}{k!}\\
&\lt 2+\sum_{k=2}^{\infty} \frac{1}{2\cdot 2^{k-2}}\\
&\lt 2+\frac1{2}\sum_{k=0}^{\infty} \frac{1}{2^k}\\
&= 2+1\\
&= 3
\end{array}
$
