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?

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 $n$.