TITLE: Combinatorial proof of a Stirling number identity.
QUESTION [29 upvotes]: Consider the identity
$$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$
where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity
$$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$
which essentially states that the number of even subsets of a set is equal to the number of odd subsets. 
Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?
As an added note, there is a similar identity
$$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$
A combinatorial interpretation of this one would also be appreciated.

REPLY [3 votes]: For the sake of completeness I include a treatment using generating functions.
The exponential generating function of the Stirling numbers of the second kind is
$$ G(z, u) = \exp(u(\exp(z)-1))$$
so that
$$ {n \brace k} = n! [z^n] \frac{(\exp(z) - 1)^k}{k!}.$$
It follows that
$$\sum_{k=0}^n (-1)^k k! {n \brace k} = n! [z^n] \sum_{k=0}^n (1-\exp(z))^k
= n! [z^n]  \sum_{k=0}^\infty (1-\exp(z))^k,$$
where the last equality occurs because the series for $(1-\exp(z))^k$ starts at degree $k.$
But this is just
$$ n! [z^n] \frac{1}{1-(1-\exp(z))} = n! [z^n] \exp(-z) = (-1)^n,$$
showing the result.