How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$

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 $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$

After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.

We will first prove the first equation. Note that $$\begin{align} 1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\vdots}{5}}{4}}{3}}{2}}{1}&=1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1}{5}+\cfrac{\vdots}{5}}{4}}{3}}{2}}{1}\\ &=1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1}{4}+\cfrac{1}{4\cdot5}+\cfrac{\vdots}{4\cdot5}}{3}}{2}}{1}\\ &=1+\cfrac{1+\cfrac{1+\cfrac{1}{3}+\cfrac{1}{3\cdot4}+\cfrac{1}{3\cdot4\cdot5}+\cfrac{\vdots}{3\cdot4\cdot5}}{2}}{1}\\ &=1+\cfrac{1+\cfrac{1}{2}+\cfrac{1}{2\cdot3}+\cfrac{1}{2\cdot3\cdot4}+\cfrac{1}{2\cdot3\cdot4\cdot5}+\cfrac{\vdots}{2\cdot3\cdot4\cdot5}}{1}\\ &=1+\cfrac{1}{1!}+\cfrac{1}{2!}+\cfrac{1}{3!}+\cfrac{1}{4!}+\cfrac{1}{5!}+\cdots\\ &=\color{red}e\tag{1} \end{align}$$ We will then proceed to the second equation. $$\begin{align} 1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5+\cfrac{6+\vdots}{5}}{4}}{3}}{2}}{1}&=1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5+\cfrac{6}{5}+\cfrac{\vdots}{5}}{4}}{3}}{2}}{1}\\ &=1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{5}{4}+\cfrac{6}{4\cdot5}+\cfrac{\vdots}{4\cdot5}}{3}}{2}}{1}\\ &=1+\cfrac{2+\cfrac{3+\cfrac{4}{3}+\cfrac{5}{3\cdot4}+\cfrac{6}{3\cdot4\cdot5}+\cfrac{\vdots}{3\cdot4\cdot5}}{2}}{1}\\ &=1+\cfrac{2+\cfrac{3}{2}+\cfrac{4}{2\cdot3}+\cfrac{5}{2\cdot3\cdot4}+\cfrac{6}{2\cdot3\cdot4\cdot5}+\cfrac{\vdots}{2\cdot3\cdot4\cdot5}}{1}\\ &=1+\cfrac{2}{1!}+\cfrac{3}{2!}+\cfrac{4}{3!}+\cfrac{5}{4!}+\cfrac{6}{5!}+\cdots\\ &=\sum_{n=1}^{\infty}\cfrac{n^2}{n!}\\ &=\color{red}{2e}\tag{2} \end{align}$$ Then, using the same logic, we have $$\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$$