Establish the given formulas for some given sums

Assume that
$\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \qquad \text{for all } \ \ x.$

Show that

$\sum_{n=1}^{\infty} \frac{n^2 x^n}{n!} = (x^2 + x)e^x,$

assuming that we can operate on infinite sums in the same way we can operate on finite sums.
Given that
$\sum_{n=1}^{\infty} \frac{n^3}{n!} = ke$

for a positive integer $k$ . Find the value of $k$ , without attempting to justify the formal manipulations of the infinite series.

We compute,
$\begin{align*} &&\sum_{n=0}^{\infty} \frac{x^n}{n!} &= e^x \\[9pt] \implies && \sum_{n=0}^{\infty} \frac{nx^{n-1}}{n!} &= e^x &(\text{differentiating})\\[9pt] \implies && \sum_{n=0}^{\infty} \frac{nx^n}{n!} &= xe^x &(\text{multiplying by }x)\\[9pt] \implies && \sum_{n=0}^{\infty} \frac{n^2 x^{n-1}}{n!} &= e^x (x+1) &(\text{differentiating})\\[9pt] \implies && \sum_{n=0}^{\infty} \frac{n^2 x^n}{n!} &= e^x (x^2+x) &(\text{multiplying by }x). \end{align*}$
Starting with the formula in part (a) (and noting that since the $n =0$ term is zero we can run the sum from $n=1$ to infinity) we have
$\begin{align*} && \sum_{n=1}^{\infty} \frac{n^2 x^n}{n!} &= e^x (x^2 + x) \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{n^3 x^{n-1}}{n!} &= e^x (x^2 + 3x + 1) \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{n^3}{n!} &= 5e &(\text{letting }x =1) \\[9pt] \implies && k &= 5. \end{align*}$

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