Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$.

Notice that \begin{align} S & = \sum_{n = 0}^{\infty} \frac{n^{2}}{n!} \\ & = \frac{0^{2}}{0!} + \frac{1^{2}}{1!} + \frac{2^{2}}{2!} + \frac{3^{2}}{3!} + \cdots \\ & = \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots. \end{align} Then \begin{align} S - e & = \left( \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots \right) - \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \right) \\ & = \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \cdots \\ & = e. \qquad (\text{As shown in the OP.}) \\ \end{align} Therefore, $ S = e + e = 2 e $.

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In fact, using the same type of reasoning, it can be shown that $ \displaystyle \sum_{n = 0}^{\infty} \frac{n^{3}}{n!} = 5 e $.

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Here is the connection with Bell numbers.

For each $ p \in \mathbb{N}_{0} $, let $ \displaystyle S_{p} \stackrel{\text{df}}{=} \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} $. Notice that \begin{align} \forall p \in \mathbb{N}: \quad S_{p} & = \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} \\ & = \sum_{n = 1}^{\infty} \frac{n^{p}}{n!} \qquad \left( \text{As $ \dfrac{0^{p}}{0!} = 0 $.} \right) \\ & = \sum_{n = 1}^{\infty} \frac{n^{p - 1}}{(n - 1)!} \qquad (\text{After canceling $ n $’s.}) \\ & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!}. \qquad (\text{After re-indexing.}) \end{align} Hence, \begin{align} \forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} - S_{p - 1} & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!} - \sum_{n = 0}^{\infty} \frac{n^{p - 1}}{n!} \\ & = \sum_{n = 0}^{\infty} \frac{1}{n!} \left[ (n + 1)^{p - 1} - n^{p - 1} \right] \\ & = \sum_{n = 0}^{\infty} \left[ \frac{1}{n!} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} n^{k} \right] \qquad (\text{By the Binomial Theorem.}) \\ & = \sum_{n = 0}^{\infty} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} \frac{n^{k}}{n!} \\ & = \sum_{k = 0}^{p - 2} \left[ \binom{p - 1}{k} \sum_{n = 0}^{\infty} \frac{n^{k}}{n!} \right] \\ & = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k}. \end{align} Therefore, $$ \forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k} + S_{p - 1} = \sum_{k = 0}^{p - 1} \binom{p - 1}{k} S_{k}. $$ As $ S_{0} = S_{1} = e $ by inspection, we see that the $ S_{p} $’s are the Bell numbers multiplied by $ e $.