# Exponential Type Series [duplicate]

I'm looking for a closed expression (if it exists) of the following sum: $$\sum_{m=0}^{\infty} \frac{m^n}{m!}c^m$$ where $$n \geq 1$$ is a positive integer, and $$c$$ is a fixed constant. The series seems to be convergent for all positive integers $$n$$, but I can't see the pattern; first values are: $$\sum_{m=0}^{\infty} \frac{m}{m!}c^m=ce^c$$ $$\sum_{m=0}^{\infty} \frac{m^2}{m!}c^m=c(c+1)e^c$$ $$\sum_{m=0}^{\infty} \frac{m^3}{m!}c^m=c(c^2+3c+1)e^c$$ But for an arbitrary $$n$$? I can't find any reference. Thanks!

## marked as duplicate by Martin R, Yanior Weg, mrtaurho, Paul Frost, Lord Shark the UnknownMay 23 at 4:08

This problem comes down to expressing $$m^n$$ as the sum of terms of form $$\prod\limits_{k=0}^{r} (m-k)$$.
Note that in the $$n=2$$ case, you had $$m^2 = m + m(m-1)$$, which yielded $$c^2 + c$$ as the coefficient of $$e^c$$. Similarly, $$m^3 = m + 3m(m-1) + m(m-1)(m-2)$$, which yielded $$c^3 + 3c^2 + c$$. In general, if you have $$m^n = \sum \limits_{r=1}^{n} \left(a_r \prod\limits_{k=0}^{r-1} (m-k) \right)$$, for integers $$a_n$$, you will obtain $$\left( \sum \limits_{r=1}^{n} a_r c^{n-r+1} \right)e^c.$$
You can determine $$a_r$$ by starting with $$m^n$$ and successively subtracting $$\prod\limits_{k=0}^{r} (m-k)$$, starting with $$r=n$$ first, and working down to $$r=1$$, ensuring to reduce the degree of the equation by $$1$$ each time.
There is not a simple closed form for these $$a_r$$ and therefore the polynomial multiplied by $$e^c$$. These polynomials are called Touchard polynomials and can be found here: https://en.wikipedia.org/wiki/Touchard_polynomials