Compute $\sum_{k=0}^n s(n,k)=\sum_{k=0}^n (-1)^{n-k}{n\brack k}$.
From definition, it's clear than the sum of the unsigned Stirling numbers is: $$\sum_{k=0}^n {n\brack k}=n!$$
But I'm stucked at $\sum_{k=0}^n s(n,k)$, since definition of the signed version is given by a falling factorial. Any hint?