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 $$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$ for $n \geq 2$. A combinatorial interpretation of this one would also be appreciated.