# Evaluating $\sum_{x=a}^{\infty}\frac{1}{e\cdot x!}{x \choose a} \cdot p^{x-a}$ for $p \in [0,1]$

I am trying to evaluate the following sum $$\sum_{x=a}^{\infty}\frac{1}{e\cdot x!}{x \choose a} \cdot p^{x-a}$$ for $$p \in [0,1]$$. This looks somewhat like the Taylor series expansion of $$e$$, but I don't know how I would go about applying it because of the binomial.

I evaluated the sum using Mathematica, and I got $$\frac{e^{p-1}}{a!}$$ which does make it seem like it's been obtained using the Taylor series of $$e$$ but I don't see how I would go about using it.

Any help would be appreciated.

$$\sum_{n=a}^{\infty} \frac{1}{n!e}\cdot\frac{n!}{a!(n-a)!} p^{n-a}= \frac{1}{a!e}\sum_{n=a}^{\infty} \frac{p^{n-a}}{(n-a)!}$$
I used $$n$$ instead of $$x$$. Notice that $$\sum_{n=a}^{\infty} \frac{p^{n-a}}{(n-a)!} = 1+p+\frac{p^2}{2!}+\dots= \sum_{k=0}^{\infty} \frac{p^k}{k!}=e^p$$ So, the final answer is $$\frac{e^{p-1}}{a!}$$.