Stirling Numbers Proof 
Let $n > 1$ be an integer. Prove the following:
  $$\sum\limits_{k=1}^{\infty} (-1)^k (k - 1)! S(n,k) = 0$$
  where $S(n,k)$ is a Stirling number of the second kind.

(Hint: Recurrence Relation)
Workings:
The recurrence relation of Stirling numbers of the second kind I believe is:
$S(n+1,k) = k S(n,k) + S(n,k-1)$
Though I do not see how this will potentially help out.
Any help will be appreciated.
 A: If you use the hint and let

$$ b_k = k! \,S(n-1,k)+(k-1)!S(n-1,k-1) = a_k+a_{k-1} $$

then the series becomes

$$ \sum\limits_{k=1}^{∞} (−1)^k b_k =  \sum\limits_{k=1}^{∞} (−1)^k (a_k+a_{k-1}) $$

and you will notice that when you expand the series terms will cancel each other.
A: Since the recursion gives:
$$ x^n = \sum_{k=1}^{n} S(n,k)(x)_k = \sum_{k=1}^{n} S(n,k)x(x-1)\cdot\ldots\cdot(x-(k-1)) $$
(see also line $(11)$ here) we have that:
$$x^{n-1}= \sum_{k=1}^{n} S(n,k)(x-1)\cdot\ldots\cdot(x-(k-1))\tag{1}$$
and the claim simply follows from evaluating the previous identity in $x=0$.
A: Seeing that the OGF of the Stirling numbers of the second kind has already been used I would like to point out that this can be done using the EGF as well. Recall the species of set partitions which is
$$\mathfrak{P}(\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which gives the bivariate generating function
$$G(z,u)  = \exp(u(\exp(z)-1)).$$
It follows immediately that
$${n\brace k} = n![z^n] \frac{(\exp(z)-1)^k}{k!}.$$
Suppose we seek to evaluate
$$\sum_{k=1}^n (-1)^k\times (k-1)! \times {n\brace k}.$$
Substitute the EGF value into the sum to get
$$n! [z^n] \sum_{k=1}^n (-1)^k \times (k-1)! 
\times \frac{(\exp(z)-1)^k}{k!}$$
which is
$$n! [z^n] \sum_{k=1}^n (-1)^k
\times \frac{(\exp(z)-1)^k}{k}.$$
Now observe that $(\exp(z)-1)^k$ starts at the power $z^k$ so we may extend the sum to infinity without affecting the count, getting
$$n! [z^n] \sum_{k\ge 1} (-1)^k
\times \frac{(\exp(z)-1)^k}{k}
= n! [z^n] \log\frac{1}{1+(\exp(z)-1)}
\\ = n! [z^n] \log \exp(-z) = n! [z^n] (-z).$$
This is $-1$ when $n=1$ and zero otherwise.
