TITLE: Prove that $(n!)^2$ is greater than $n^n$ for all values of n greater than 2.
QUESTION [5 upvotes]: This problem , I assume can be proved using induction, however I am trying to find another way. 
Is there a simple combinatorial approach? One notices that $(n!)^2$ is equal to the number of permutations of size n squared, and that $n^n$ is the number of redundant combinations where there are n spaces and n choices. 
Any help would be much appreciated.
Thanks

REPLY [3 votes]: This is not combinatorial, but note that
$$(n!)^2=\prod_{k=1}^n k(n+1-k).$$
(We are in essence using the "Baby Gauss" trick.)
But if $k$ is not $1$ or $n$, we have $k(n+1-k)\gt n$.