Sum involving Stirling's numbers of the second kind

Question

I was reading one book about basic probability theory and came across one problem that caught my attention. Problem description
So in the end there is a sum of the form: $\sum_{i=0}^{n} (-1)^{i}\binom{n}{i}(1-\frac{i}{n})^k$.
The author then says that in fact, this is just $\frac{n!}{n^k}S_{n,k}$, where $S_{n,k}$ is Stirling numbers of the second kind. I don't understand how he simplified the sum. Please help.
P.S. Sorry if the question is a bit stupid.

Answer 1

You can rewrite this as $$n! S_{n,k}=\sum_{i=0}^n(-1)^i\binom{n}i(n-i)^k.\tag{*}$$ This is a slightly eccentric notation for the Stirling numbers, as we have to interpret $S_{n,k}$ as the number of partitions of $[k]=\{1,\ldots,k\}$ into $n$ non-empty parts.

In (*) both sides count the number of surjections from $[k]$ to $[n]$. The LHS is pretty much by definition, the RHS uses inclusion-exclusion.