$$\begin{align}\\ S(n,r) &= \frac1{r!}\sum_{i = 0}^r (-1)^i\binom r i(r-i)^n\\ &= \frac1{r!}\sum_{i = 0}^r (-1)^{r-i}\binom r i i^n\\ \end{align}$$
I know how to derive the formula in the first line using inclusion-exclusion, but:
How did we go from the first line to the second?
We have
$$S(n,r)=\frac1{r!}\sum_{i = 0}^r (-1)^i\binom{r}{i}(r-i)^n$$
Note that $i$ runs from $0$ to $r$, but we can also run from $r$ to $0$, which is equivalent with replacing all $i$'s with $r-i$. We get
$$S(n,r)=\frac1{r!}\sum_{i = 0}^r (-1)^{r-i}\binom{r}{r-i}(r-(r-i))^n$$
Note that $\binom{r}{r-i}=\binom{r}{i}$ so that this reduces to
$$S(n,r)=\frac1{r!}\sum_{i = 0}^r (-1)^{r-i}\binom{r}{i}i^n$$
en that's the last line.