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 $\infty$.
 A: The easy part first: We have by basic analysis (where $x\in\Bbb R$)
\begin{equation}\tag 1\label 1\lim_{x\to\infty} \left(\frac{x+1}x\right)^x=\lim_{x\to\infty}(1+1/x)^x=e.\end{equation}
Now comes the harder part:
Note that
\begin{equation}\label 2\tag 2
\lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}
= \lim_{x\to\infty} \exp\left(x+x^2\ln\left({x+1\over x+2}\right)\right).
\end{equation}
We now have by Taylor expansion of $\ln(1-y)$ (for $x$ big enough):
\begin{align}\tag 3\label 3
x+x^2
\ln\left(1-\frac1{x+2}\right)
&=x-\sum_{k=1}^\infty\frac1k\frac{x^2}{(x+2)^k}
\\ &=x-\frac{x^2}{x+2}-\frac{x^2}{2(x^2+4x+4)}-\overbrace{\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}}^{\xrightarrow{x\to\infty} 0}.
\end{align}
The latter sum converges to $0$ since (for $x> 1$), $$\displaystyle\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}\le\sum_{k=3}^\infty \frac{x^2}{x^k}=\sum_{k=1}^\infty x^{-k}=\frac1x\frac{x}{x-1}=\frac1{x-1}.$$
Thus, by additivity of the limit,
\begin{align}\label 4\tag 4
\lim_{x\to\infty} x+x^2\ln(1-\frac1{x+2}) = \lim_{x\to\infty} \overbrace{x-\frac{x^2}{x+2}}^2-\overbrace{\frac{x^2}{2(x^2+4x+4)}}^\frac12=\frac32.
\end{align}
We can now use continuity of the exponential function and \eqref{2} to find that
\begin{align}\tag 5\label 5
\lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}
&= \exp\left(\lim_{x\to\infty} x+x^2\ln\left({x+1\over x+2}\right)\right) \\
&= e^{3/2}.&\eqref 4
\end{align}
We can thus finally assert, using multiplicity of the limit, that your limit equals $0$:
\begin{equation}
\bbox[5px,border:2px solid #C0A000]{
\lim_{x\to\infty} \color{orange}{\frac{x+1}{(x+2)\sqrt x}}
\color{blue}{e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}}
\color{green}{\left(\frac{x+1}x\right)^x}
= \color{orange}0 \cdot \color{blue}{e^{3/2}} \cdot \color{green}e = 0.
}
\end{equation}
A: $$\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$$
= $$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^{n^2+n+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^nn^{n^2+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} n^n$$
=$$\frac{((1+\frac{1}{n})^{n})^n(1+\frac{1}{n})^{n}(1+\frac{1}{n})}{n! ((1+\frac{2}{n})^{n})^n (1+\frac{2}{n}) } n^n$$
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
