stirling number relation $(xD)^n = \sum_{k \in \mathbb Z} S(n,k) x^k D^k$

Question

Prove the stirling number of second kind relation where $D$ is the differential operator $$(xD)^n = \sum_{k \in \mathbb Z} S(n,k) x^k D^k$$

Not sure how to do this, please help!

Answer 1

You can use induction on $n$.

Alternatively you can verify that both sides have the same effect on $x^m$. This boils down to the identity $$m^n=\sum_k S(n,k)k!\binom{m}k$$ which has a combinatorial proof: count the number of maps from an $n$-element set to an $m$-element set according to the size of image.