Inequality $(n!)^2\le \left[\frac{(n+1)(n+2)}{6}\right]^n$ Prove that
$$
(n!)^2\le \left[\frac{(n+1)(n+2)}{6}\right]^n
$$
holds for all $n\in\mathbb{Z^+}$.
I tried induction but there's no obvious way to go from $n$ to $n+1$. 

Try to prove first that
$$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6.$$
(I have posted a separate question about this sum.)
Then use the AM-GM inequality for the sum on the LHS.
