Show $\left( n!\right)^2 > n^n$. If $n > 2$, show that $$\left(n!\right)^2 > n^n$$
Although the problem is pretty obvious, I couldn't come up with a rigorous proof. I was thinking some sort of AM-GM, but couldn't build anything concrete.
I have a feeling this problem could be a duplicate, I tried to google, and even searched on the stackexchange network, couldn't find it.
 A: Base case, $n=3$, is true by $(3!)^2 = 36 > 27 = 3^3$.
Suppose that we have for $k>2$, $(k!)^2 > k^k$ (IH), now we want to show that $((k+1)!)^2 > (k+1)^{k+1}$.
\begin{align*}
((k+1)!)^2 &= ((k+1)(k!))^2\\
&=(k+1)^2(k!)^2>(k+1)^2k^k \qquad\text{   (where the inequality is from IH)}
\end{align*}
So if we can show that $(k+1)^2k^k>(k+1)^{k+1}$ then we are done. I'll leave that up to you.
A: We do the induction step. If $(k!)^2\gt k^k$, and $k\ge 3$, we show that $((k+1)!)^2 \gt (k+1)^{k+1}$, Note that $((k+1)!)^2 =(k+1)^2(k!)^2 = (k+1)^2(k!)^2 \gt (k+1)^2 k^k$.
We show that $(k+1)^2 k^k \gt (k+1)^{k+1}$, or equivalently that $(k+1)k^k \gt (k+1)^k$. So we need to show that $k+1 \gt \left(1+\frac{1}{k}\right)^k$. This is true, since $\left(1+\frac{1}{k}\right)^k \lt e$.  
Another way: Note that $(n!)^2 =\prod_1^n k(n+1-k)$. The function $x(n+1-x)$ increases until $x=\frac{n+1}{2}$, then decreases. In particular, $k(n+1-k)\ge (1)(n)=n$. So the product $\prod_1^n k(n+1-k)$ is $\ge$ the product of $n$ copies of $n$. 
A: $$
n\ge k\ge  1\implies n(k-1)\ge k^2-k\implies  nk-k^2+k\ge n
\\
\implies n-k+1\ge \frac nk 
$$
Now make the product of these inequalities for $1\le k \le n$:
$$n! = (n-1+1)(n-2+1)\cdots (n-n+1)
\ge \frac {n^n}{n!}
$$
The macroscopic inequality is strict as soon as it is strict for one microscopic inequality, 
that is when there is $k$ such as $1<k<n$, ie $n>2$.
A: It's easy to see that
$${n\choose k}\left({1\over n}\right)^k\lt{n\over n}\cdot{n-1\over n}\cdot\cdots\cdot{n-k+1\over n}\lt1$$
if $n\ge2$ and hence
$$\left(1+{1\over n}\right)^n=\sum_{k=0}^n{n\choose k}\left({1\over n}\right)^k\lt\sum_{k=0}^n1=n+1$$
It now follows that
$$(n+1)^{n+1}=(n+1)\left(1+{1\over n}\right)^nn^n\lt(n+1)^2n^n\le(n+1)^2(n!)^2=((n+1)!)^2$$
using induction (and the base case $2^2=(2!)^2$) for the final inequality.
