Through some calculation, it can be shown that $$e = 1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{1}$$ $$2e = 1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{2}$$ $$5e = 1+\cfrac{2^2+\cfrac{3^2+\cfrac{4^2+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{3}$$ In general, how can I show that
$$\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$$ , where $B_n$ is the $n^{th}$ Bell number.
I saw a similar question on Brilliant.org, but I did not pay close attention to the proof, and I ended up forgetting how to prove this kind of problem. I remember that the proof involves in simplifying the denominator and moving from top to bottom so that the final denominator is in the form of $n!$, which is the criteria for Maclaurin series.
Here is the background information Infinite Series $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$
