How to calculate limit as $n$ tends to infinity of $\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$? This question stems from and old revision of this question, in which an upper bound for $n!$ was asked for. 
The original bound was incorrect. In fact, I want to show that the given expression divided by $n!$ goes to $0$ as $n$ tends to $\infty$.

I thus want to show:
  $$\lim_{n\to\infty}\frac{(n+1)^{n^2+n+1}}{n!(n+2)^{n^2+1}}=0.$$

Using Stirling's approximation, I found that this is equivalent to showing that
$$\lim_{n\to\infty} \frac{\exp(n)}{\sqrt n}\cdot\left(\frac{n+1}{n+2}\right)^{n^2+1}\cdot\left(\frac{n+1}{n}\right)^n=0.$$
However, I don't see how to prove the latter equation.
EDIT: It would already be enough to determine the limit of $$\exp(n)\left(\frac{n+1}{n+2}\right)^{(n^2)}\left(\frac{n+1}{n}\right)^n$$ as $n$ goes to $\infty$.

$$\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$$
= $$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^{n^2+n+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^nn^{n^2+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} n^n$$
=$$\frac{((1+\frac{1}{n})^{n})^n(1+\frac{1}{n})^{n}(1+\frac{1}{n})}{n! ((1+\frac{2}{n})^{n})^n (1+\frac{2}{n}) } n^n$$
at the limit n tends to infinity, using the standard definition of log(z) = $\lim_{x\to\infty}(1 + 1/z)^z$
=$$\frac{e^n.e.1}{n!e^{2n}e^2} n^n$$
=$$\frac{ n^n}{n!e^{n+2}} $$
is how far I got really
