<s>[INST] How to Prove : $\frac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\frac{n(3n+1)}{12}$ While I calculate an integral 
$$
\int\limits_{[0,1]^n}\cdots\int(x_1+\cdots+x_n)^2\mathrm dx_1\cdots\mathrm dx_n
$$
I used two different methods and got two answers. I am sure it's equivalent, but how can I prove it?
$$\displaystyle\dfrac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\dfrac{n(3n+1)}{12}$$
Sincerely thanks!
 [/INST] Fix $n$ and define $f(x)$ by
$$ f(x) = (1 - e^{-x})^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k e^{-kx}. $$
Then the left-hand side is equal to
$$ \frac{2}{(n+2)!} f^{(n+2)}(0). $$
So it suffices to find the Taylor polynomial of degree $n+2$ for $f(x)$ at $x = 0$. This can be done as follows:
\begin{align*}
f(x)
&= x^n \left( \frac{1 - e^{-x}}{x} \right)^n \\
&= x^n \exp\left[ n \log\left( \frac{1 - e^{-x}}{x} \right)\right] \\
&= x^n \exp\left( -\frac{n}{2} x + \frac{n}{24} x^2 + \cdots \right) \\
&= x^n \left( 1 - \frac{n}{2} x + \frac{(3n+1)n}{24} x^2 + \cdots \right) \\
&= x^n - \frac{n}{2} x^{n+1} + \frac{(3n+1)n}{24} x^{n+2} + \cdots.
\end{align*}
Therefore we have
$$ \frac{2}{(n+2)!} f^{(n+2)}(0) = \frac{2}{(n+2)!} \cdot \frac{(3n+1)n}{24} (n+2)! = \frac{(3n+1)n}{12}. $$
</s>