Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ ?
$$S_3 = \sum_{n \geq 0} \frac{n^3}{n!}=\sum_{n \geq 1} \frac{n^2}{(n-1)!} \implies$$
$$S'_3=S_3-e=\sum_{n \geq 1} \frac{n^2-1^2}{(n-1)!}=\sum_{n \geq 2} \frac{n+1}{(n-2)!} \implies$$
$$S'_3-3e=\sum_{n \geq 2} \frac{(n+1)-3}{(n-2)!}=\sum_{n \geq 3} \frac{1}{(n-3)!}=e\implies$$
$$S'_3-3e=e\iff S_3=5e$$
Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$. : One of the answers shows how $\sum_{n \geq 0} \frac{n^2}{n!}=2e$ and asserts that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ can be showed in the same manner of reasoning. Is my "proof" correct?

Seems good to me. An alternate way - which is slightly neater - would be to first write $n^3$ in a convenient way:
\begin{align} n^3 &= n^2(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n(n-1) + n.\end{align}
Then, since all the sums involved converge, \begin{align} \sum_{n \ge 0} \frac{n^3}{n!} &= \sum_{n \ge 0} \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!} \\ &= \sum_{n\ge 3} \frac{1}{(n-3)!} + \sum_{n \ge 2} \frac{3}{(n-2)!} + \sum_{n \ge 1} \frac{1}{(n-1)!} = 5e.\end{align}
Expressions such as $n(n-1)(n-2)$ are called falling factorials:
$(n)_k = n (n-1) \dots(n-k+1)$. These are sometimes also denoted $n^{\underline{k}}$. There is a standard way to write powers as falling factorials: $$ n^k = \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix} (n)_i, $$ where $\begin{Bmatrix} n \\ i\end{Bmatrix}$ are the Stirling numbers of the second kind. Using this, and the observation that $\frac{(n)_i}{n!} = \frac{1}{(n-i)!},$ you can generalise the above decomposition to any $k$ instead of $3$, and derive a similar result, that $$ \sum_{n \ge 0} \frac{n^k}{n!} = \left( \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix}\right) e = B_k e,$$ where $B_k$ are the Bell numbers, that are simply the sum in the parenthesis above.
