Prove that for all positive integers $n$, $\sum_{k=1}^{n}(-1)^kk!S(n,k)=(-1)^n$

Question

Prove that for all positive integers $n$, $$\sum_{k=1}^{n}(-1)^kk!S(n,k)=(-1)^n$$ where $S(n,k)$ are the Stirling numbers of the second kind. I recognize that $k!S(n,k)$ is the number of surjections from an n-element space to a k-element space. I think there might be some kind of symmetry as well, but I am unsure how to evolve this thought into the proof. For instance, $$k!S(n,k) = 1 \text{ for } k=1 \text{ and }k=n$$

Answer 1

Here is an interesting idea:

We know the following:

$$k!S(n,k)=\sum_{i=0}^{k}(-1)^i\binom{k}{i}(k-i)^n$$

so we want to prove

$$\sum_{k=1}^{n}\bigg((-1)^k\sum_{i=0}^{k}(-1)^i\binom{k}{i}(k-i)^n\bigg)=(-1)^n$$

but we observe that

$$\sum_{k=1}^{n}\bigg((-1)^k\sum_{i=0}^{k}(-1)^i\binom{k}{i}(k-i)^n\bigg)=\sum_{i=1}^{n}i^n(-1)^i\bigg(\binom{i}{0}+\binom{i+1}{1}+...+\binom{n}{n-i}\bigg)$$