Proving $\sum_{k=0}^\infty x^kk^n/k!=(x\, \mathrm d/\mathrm dx)^n e^x$ On this tweet the following equality is shown:
$$\sum_{k=0}^\infty \left( \frac{x^k}{k!}  \right)k^n=\left( x \frac{\mathrm d}{\mathrm dx}\right)^n e^x$$
I can see that on the RHS the derivative of $e^x$ is always $e^x,$ and I don't know whether the idea symbolized in there is to take $n$ consecutive derivatives, or the effect of keeping the $x$ within the parenthesis in $\left( x\frac{\mathrm d}{\mathrm dx} \right)^n.$
On the LHS I see the expansion of $e^x= 1+x+\frac{e^2}{2!}+\cdots,$ but the $k^n$ is a mystery.

What are the missing pieces to understand this equality?

 A: The $x \frac{\mathrm{d}}{\mathrm{d}x}$ is a linear map, which we will call $T$. We have, for example
\begin{align*}
T(e^x) &= xe^x \\
T(\log x) &= 1 \\
T(x^n) &= nx^n \\
T(\cos(x)) &= -x\sin(x),
\end{align*}
etc. The question is, if you apply $T$ to $e^x$ $n$ times, do you obtain the function represented by the given power series?
The clear solution in my mind is induction on $n$. When $n = 0$, we apply $T$ to $e^x$ $0$ times, leaving just $e^x$ on the left side. On the right side, we get the usual series for $e^x$, thus our base case is true.
Now, assume that
$$\sum_{k=0}^\infty \frac{x^k}{k!}k^n= T^n (e^x)$$
for some fixed $n$. Then,
\begin{align*}
T^{n+1}(e^x) &= T(T^n(e^x)) \\
&= T\left( \sum_{k=0}^\infty \frac{x^k}{k!} k^n\right) \\
&= x\sum_{k=0}^\infty k\frac{x^{k-1}}{k!} k^n \\
&= \sum_{k=0}^\infty \frac{x^k}{k!} k^{n+1},
\end{align*}
as required. Thus the result holds, by induction, for all $n$.
A: First, note that
$$
x\frac d{dx} x^k = nx^k.
$$
Repeating, it follows that
$$
\left(x\frac{d}{dx}\right)^n x^k = k^nx^k.
$$
Now, ignoring convergence issues (there aren't any),
$$
\begin{aligned}
\left(x\frac{d}{dx}\right)^n e^x
&= \left(x\frac{d}{dx}\right)^n \sum_{k=0}^\infty \frac{x^k}{k!}\\
&= \sum_{k=0}^\infty \frac1{k!}\left(x\frac{d}{dx}\right)^n x^k\\
&= \sum_{k=0}^\infty \frac1{k!}k^nx^k\\
&= \sum_{k=0}^\infty \frac{x^k}{k!}k^n.
\end{aligned}
$$
