Calculation of sum of infinite products $\sum_{n=1}^\infty n*\frac{r^n}{n!}$ Is there any analytical solution to this sum of infinite products??
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!}$$
and
$$\sum_{n=1}^\infty n^2\cdot \frac{r^n}{n!}$$

Hint for $1$:
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!} = r \sum_{n=1}^\infty  \frac{r^{n-1}}{(n-1)!} = r \sum_{n=0}^\infty \frac{r^n}{n!}$$
Hint for $2$:
$$\sum_{n=1}^\infty n^2\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty (n(n-1) + n)\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty n(n-1)\cdot\frac{r^n}{n!}+\sum_{n=1}^\infty n\cdot\frac{r^n}{n!}$$
And
$$\sum_{n=1}^\infty n(n-1)\cdot\frac{r^n}{n!}=r^2\sum_{n=1}^\infty n(n-1)\cdot\frac{r^{n-2}}
{n!} \tag{1}$$
You get $(1)$ by differentiating $2$ times w.r.t. $r$ and then multiplying by $r^2$.
$$\sum_{n=0}^\infty \frac{r^n}{n!}$$
