Is this Inclusion-Exclusion: $m$ balls colored $n$ colors such that each color is represented? Suppose you have $m$ balls each of which can be colored by $n<m$ different colors, and your paint cans are unlimited.  I want to count the number of ways that you can color balls such that each color is represented at least once.  I reasoned that this can be counted by something that seems ... "philosophically" similar to inclusion-exclusion.  But when I try to actually align this counting method with the actual inclusion-exclusion method, I can't seem to perfectly describe how this counting method is an application of inclusion-exclusion.
Here's my solution:  First over-estimate the number of colorings by all possible assignments of colors to balls:  $n^m$.
Next we subtract out those colorings in which some color is absent.  There are $\binom n 1$ choices of one color, and for each, we can color each ball by $(n-1)^m$ colors.  So our new estimate is
$$ n^m - \binom n 1 (n-1)^m $$
However, there are some colorings that are double-counted by this subtraction.  In particular the colorings which are lacking two colors are double-counted (and the colorings missing three colors are triple-counted, and so on).  So to correct for this we add back in $\binom n 2 (n-2)^m$.
$$ n^m - \binom n 1 (n-1)^m + \binom n 2 (n-2)^m $$
We continue the process until, with just one color left, double-countings no longer occur.  So the number is
$$ \sum_{i=0}^{n-1} (-1)^i \binom{n}{n-i}(n-i)^m $$
Edit: I tried searching for this in the recommendations that StackExchange gives, and didn't see anything that looked very relevant.  But now on the sidebar I see this link:  Coloring m things with k colors, using each color at least once.
So this is a known problem, but is there a way to view it as an instance of inclusion-exclusion, or is the similarity just superficial?

I would leave it as
$$\sum_{i=0}^{n-1}(-1)^i\binom{n}i(n-i)^m$$
or even
$$\sum_{i=0}^n(-1)^i\binom{n}i(n-i)^m\,,$$
but what you’ve done is fine and is exactly an inclusion-exclusion argument. The Stirling numbers mentioned in the answer to the earlier question that you found can be calculated by essentially the same inclusion-exclusion argument.
