Show that $n! < \frac{(n+1)^{n^2+n+1}}{(n+2)^{n^2+1}}$ for $n\ge 2$

Question

$n! < \frac{(n+1)^{n^2+n+1}}{(n+2)^{n^2+1}}$

For very small values of $n$ (i.e. $2\le n\le 6$) the function on the right nicely approximates $n!$ before significantly overtaking it. I don't have much work to show (unfortunately) because this is just what my calculations came down to after a rather long proof. That is to say, proving this will close up the rest of my proof.

Unfortunately, induction doesn't seem to be very helpful since (if we assume the induction hypothesis then):

$$(n+1)!=(n+1)n! < \frac{(n+1)^{n^2+n+2}}{(n+2)^{n^2+1}} \not< \frac{(n+2)^{n^2+3n+2}}{(n+3)^{n^2+2n+2}}$$

Answer 1

A small note on your bound: In this answer, I showed that the suggested upper bound divided by $n!$ actually converges to $0$ as $n\to\infty$. So your bound is actually a lower bound of $n!$ for large $n$.

Answer 2

If we apply Stirling approximation and Taylor expansions to $$A_n=\log (n!)-\left(n^2+n+1\right) \log (n+1)+\left(n^2+1\right) \log (n+2)$$ $$A_n==\frac{1}{2} (\log (2 \pi n)-5)+\frac{47}{12 n}+O\left(\frac{1}{n^2}\right)$$