You should start with the case where $p(i)$ is a monomial of the form $i^n$ and get the general results via linearity. In the monomial case the series is $$\sum_{i \geq 1} \frac{i^n}{i!} $$ In order to evalute this sum consider the function $$f_n(x)= \sum_{i \geq 1} \frac{x^i}{i!}=e^x-1. $$ If we differentiate both sides WRT $x$ we obtain $$e^x=\sum_{i \geq 1} \frac{i x^{i-1}}{i!} $$ Next, multiply both sides by $x$ to find $$x e^x=\sum_{i \geq 1} \frac{i x^{i}}{i!} $$ Plugging in $x=1$ we find the first sum: $\sum_{i \geq 1} \frac{i}{i!}=e$.
In order to find the sum with higher powers, consider the differential operator $\Theta=x \frac{\mathrm{d}}{\mathrm{d} x}$ (differentiation followed by multiplication by $x$). It follows that $$\sum_{i \geq 1} \frac{i^n}{i!}=\Theta^n \left[e^x-1 \right] \bigg|_{x=1}. $$ (Here $\Theta^n$ represents applying the operator $\Theta$ $n$ consecutive times.)