The sequence is:$$ a_n=\sum_{k\ge0}(k+1) {n\brack {k+2}}\;B_k\ n^k$$ $B_k$ is the Bernoulli number. ${n\brack {k}}\;$ is the unsigned Stirling number of first kind, $\left( {0\brack {0}}\;=1 \text{ and }{{n}\brack {k}}\;=(n-1){{n-1}\brack {k}}\;+{{n-1}\brack {k-1}}\;\right)$.
From $n=0$, the first terms are: $ \ \ 0\ ,\ 0\ ,\ 1\ ,\ 0\ ,\ -5\ ,\ 0\ ,\ 238\ ,\ 0\ ,\ -51508\ ,\ 0\ ,\ 35028576\ , ..$
They seem to be all integers, and the odd-indexed seem to vanish. The generating function should be even, but I could not find it. Any possible combinatorial interpretation (when removing the sign)?
EDIT(i): I have now a proof that these numbers are integers.
EDIT(ii):
I don't know for what these numbers would be relevant, nor whether they are of any importance in mathematics. I stumbled on them while playing with an extended congruence modulo arbitrary high powers of an odd prime $p$ for the generalized Harmonic numbers $H^{(j)}_{p-1}$ $\ $ $\big(H^{(m)}_{n}=\sum_{j=1}^{n}\frac{1}{j^m} \big)$, involving the Bernoulli numbers $B_j$. (I have a proof for it): \begin{align*}\sum_{j=0}^{n}B_j H^{(j+1)}_{p-1}(-p)^j& \equiv 0 \pmod{p^{n+1}} \end{align*}
I was curious to see what would happen with $$G^{(j+1)}_{p-1}=\sum_{1\le i_1< i_2<...<i_{j+1}\le p-1}\frac{1}{i_1\cdot i_2\cdot \cdot \cdot i_{j+1}}= \frac{{p\brack {j+2}}}{(p-1)!}$$ instead of $$ H^{(j+1)}_{p-1}=\sum_{i=1}^{p-1}\frac{1}{i^{j+1}}$$ since it is known that the $H^{(m)}_{n}$ and $G^{(m)}_{n}$ are interelated (there is a recurrence relation involving both kinds of numbers). Then using the first instances of this recurrence, I was able to guess - but not to prove :-( that $$\sum_{j\ge0}(j+1)B_j G^{(j+1)}_{p-1}p^j=0 $$
And then, while checking numerically, I found out, to my surprise, that $p$ prime was not needed, and that this was apparently (numerically) true under the less restrictive condition that $p$ be odd.