Finite sum related to Stirling numbers I am wondering if there is a closed form solution for the following sum:
$$
\sum _{k =0}^{n-1} \frac{(-1)^{k} (n-k)^{n+1} }{(k+1)(k+2)}\binom{n}{k}.
$$
If the the factors $(k+1)(k+2)$ in the denominator weren't there, the sum would equal $n!S(n+1,n)$, where $S(n,m)$ denotes a Stirling number of the second kind. This made me wonder if 'Stirling polynomials' of the form
$$
\mathcal{S}(x)=\sum _{k =0}^{n-1} (n-k)^{n+1}\binom{n}{k}x^k
$$
admit a closed form solution. The value of my sum could then be obtained by integrating $\mathcal{S}$ twice.
Any other strategy to evaluate the sum are also very welcome.

We can get a better closed form using a different technique.
Suppose we seek to evaluate
$$\sum_{k=0}^{n-1} \frac{(-1)^k(n-k)^{n+1}}{(k+1)(k+2)}
{n\choose k}.$$
This is
$$\frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^k (n-k)^{n+1}
{n+2\choose k+2}
\\ = \frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^{k+2} (n+2-(k+2))^{n+1}
{n+2\choose k+2}
\\ = \frac{1}{(n+2)(n+1)}
\sum_{k=2}^{n+1} (-1)^{k} (n+2-k)^{n+1}
{n+2\choose k}
\\ = -\frac{(n+2)^n}{n+1}
+ (n+1)^n +
\frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n+2} (-1)^{k} (n+2-k)^{n+1}
{n+2\choose k}.$$
Restrict to the sum for a moment and introduce
$$(n+2-k)^{n+1}
= \frac{(n+1)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2-k)z) \; dz.$$
Observe that this is zero when $k=n+2$ which is the correct value.
We get for the sum
$$\frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2)z) 
\sum_{k=0}^{n+2} (-1)^{k}
{n+2\choose k} \exp(-kz) \; dz
\\ = \frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2)z)
(1-\exp(-z))^{n+2} \; dz
\\ = \frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} 
(\exp(z)-1)^{n+2} \; dz$$
This is
$$\frac{n!}{(n+2)} [z^{n+1}] (\exp(z)-1)^{n+2} = 0$$
because $\exp(z)-1 = z +\frac{1}{2} z^2 + \frac{1}{6} z^3+\cdots.$
We conclude that the initial sum is equal to
$$-\frac{(n+2)^n}{n+1} + (n+1)^n.$$
