Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$

Alternatively, let $E=\sum \frac{z^n}{n!}=e^z$. Then

$A=zE'=\sum \frac{nz^n}{n!}=ze^z$

$B=zA'=\sum \frac{n^2z^n}{n!}=(z^2+z)e^z$

$C=zB'=\sum \frac{n^3z^n}{n!}=(z^3+3z^2+z)e^z$

Therefore, $C(1)=5e$.

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.