It seems obvious that:
$$n^n \in O(n!^2)$$
But I can't seem to find a good way to prove it.
Use a multiplicative variant of Gauss's trick:
$$
(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots
                ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)
                \ge n^n
$$