Prove that $(n!)^2$ is greater than $n^n$ for all values of n greater than 2. 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

divide $(n!)^2 > n^n$ by $n!$ to get
$$n! = 1 \times 2 \times \ldots \times (n-1)  \times n > \frac{n}{n} \times \frac{n}{n-1} \times \ldots \times \frac{n}{2} \times \frac{n}{1}$$
It is a bit of simple algebraic manipulation to show that each term on the lhs is greater or equal to the corresponding term on the rhs, or $\frac{n}{n-k} < k+1 $ for $k \in \overline{0, n-1}$
