How are the Bell numbers related to this exponential series?

$\newcommand\D{\text{D}}$ $\newcommand\Stir[2]{ {#1 \brace #2} }$ $\newcommand\diff[2]{\frac{\text{d} #1}{\text{d} #2}}$It is well known that stirling numbers of the second kind $\smash{\Stir{a}{b}}$ are related to the operator $\smash{x\D\equiv x\diff{}{x}}$

$$(x\D)^k\equiv\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\D^j\tag{1}\label{1}$$

Which can be confirmed by checking that the coefficients of $x^{k-j}\D^j$ obey the recurrence relation for Stirling numbers of the second kind.

Then operating $\eqref{1}$ on $e^x$ we have, since $\smash{\D^j(e^x)}=e^x$

$$(x\D)^ke^x= e^x\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\tag{2}\label{2}$$

by writing $\smash{e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}}$the left hand side of $\eqref{2}$ is

$$(x\D)^ke^x=\sum_{n=0}^{\infty}\frac{n^k}{n!}x^n$$

therefore

$$\sum_{n=0}^{\infty}\frac{n^k}{n!}x^n=e^x\sum_{j=0}^{k}\Stir{k}{j}x^{k-j}\tag{3}\label{3}$$

so putting $x=1$ in $\eqref{3}$ gives

$$\sum_{n=0}^{\infty}\frac{n^k}{n!}=e\sum_{j=0}^{k}\Stir{k}{j}\tag{4}\label{4}$$

then because the $n^{\text{th}}$ Bell number $B_n$ is given by

$$B_n=\sum_{j=0}^{k}\Stir{k}{j}\tag{5}\label{5}$$

we have your relation by substituting $\eqref{5}$ in to $\eqref{4}$:

$$ \sum_{n=0}^{\infty}\frac{n^k}{n!}=eB_n\tag{6}\label{6}$$