How would one prove this combinatorically?
Trying to think of some other example where this looks more intuitive. note that n! is also S(n+1,1), and there's also this recursive relationship S(n+1,k)=nS(n,k)+S(n,k-1), not sure if that really helps. Also note that sterling numbers of the first kind can be alternatively generated by this generating function.
The (unsinged) Sterling number of the first kind $\left[^n _k \right]$ are defined as the number of elements of $S_n$ (the symmetric group on $n$ elements) consisting of $k$ orbits. Recall that an element of $S_n$ can be expressed as \begin{eqnarray*} (a_1 \cdots a_{i_1}) \cdots (k_1 \cdots k_{i_k}) \end{eqnarray*} The number of brackets $k$ is the number of orbits. Thus we are just counting the elements of $S_n$ and grading/grouping them according to the number of orbits. So \begin{eqnarray*} \sum_{k=0}^n \left[^n _k \right] =n! \end{eqnarray*} $\left[^n _0 \right]=0$ is worth noting.