Proof of $\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}r^n = n!$

Question

I want to provide proof for the following equation:

$$\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}r^n = n!$$

I think it resembles the typical way of representing the Principle of Exclusion & Inclusion:

$$\sum_{\emptyset \ne J \in [n]}(-1)^{\mid J\mid-1}\mid\cap_{j\in J}Z_j\mid$$

However, still vague to let $Z_j$ to be which set to connect two different representations.

any advice or pinch of hints?