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!

Suppose we  seek to evaluate
$$\sum_{k=0}^n (-1)^k {n\choose k} (n-k)^{n+2}.$$
We introduce
$$(n-k)^{n+2} =
\frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp((n-k)z) \; dz.$$
This yields for the sum
$$\frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp(nz)
\sum_{k=0}^n (-1)^k {n\choose k} \exp(-kz) \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp(nz)
(1-\exp(-z))^n \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}}
(\exp(z)-1)^n \; dz.$$
This is
$$(n+2)! [z^{n+2}] (\exp(z)-1)^n.$$
Observe that $\exp(z)-1$ starts with $z + \frac{1}{2} z^2 + \frac{1}{6} z^3 +\cdots.$
There  are two possibilities here depending on which term from the series expansion is included in the product of the $n$ terms, the  first is  a single  term with $z^3$ and the rest being $z$ which yields
$${n\choose 1} 1^{n-1} \frac{1}{6} = \frac{1}{6} n.$$
The other possibility is two terms in $z^2$ with the rest being $z$
which yields
$${n\choose 2} 1^{n-2} \frac{1}{2^2} = \frac{1}{8} n (n-1).$$
Adding these we obtain
$$\frac{n(3n+1)}{24}$$
for a final answer of
$$(n+2)! \frac{n(3n+1)}{24}.$$
