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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?

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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 $n$.

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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), $$ \begin{aligned} \left(x\frac{d}{dx}\right)^n e^x &= \left(x\frac{d}{dx}\right)^n \sum_{k=0}^\infty \frac{x^k}{k!}\\ &= \sum_{k=0}^\infty \frac1{k!}\left(x\frac{d}{dx}\right)^n x^k\\ &= \sum_{k=0}^\infty \frac1{k!}k^nx^k\\ &= \sum_{k=0}^\infty \frac{x^k}{k!}k^n. \end{aligned} $$

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