Infinite sum involving powers and factorials I am interested in evaluating the following infinite sum
\begin{equation}
\sum_{n=0}^{\infty} \frac{\alpha^{n}}{n!}n^{\beta}
\end{equation}
where both $\alpha$ and $\beta$ are real numbers. However, in addition, $\alpha$ is always positive.
Clearly the sum converges for any value of $\alpha$ and $\beta$ since the factorial kills both exponential and power terms for sufficiently large $m$'s.
Does the sum have a closed form?
 A: Assuming $\alpha=e^{\gamma}$,
$$\sum_{n\geq 0}\frac{\alpha^n}{n!}n^\beta = \sum_{n\geq 0}\frac{e^{\gamma n}}{n!}n^{\beta}=\left.\frac{\partial^\beta}{\partial u^{\beta}}\,\exp(\exp( u))\right|_{u=\gamma}.$$
If $\beta\not\in\mathbb{N}$, see fractional calculus.
A: Indeed, it is likely that anything approaching a closed form can be only achieved for $b=0,1,2,3,\dots$
Here's some additional information not provided in the answer (though touched upon in  uranix's comment).
Writing the series as:
$$f(a,b)=a \sum_{n=0}^{\infty} \frac{(n+1)^b}{(n+1)!} a^n$$
We consider the ratio of successive terms and find by definition that for $b=1,2,3,4,5,\dots$ the series equals to a generalized hypergeometric function:

$$f(a,b)=a \cdot~{_{b-1} F_{b-1} } (\mathbf{2};\mathbf{1};a)$$

Where $\mathbf{2}=2,2,\ldots,2$ repeated $b-1$ times.
Fortunately, these functions can be expressed as polynomials (which can also be directly obtained by taking derivatives as shown by uranix and Jack D'Aurizio):
$$f(a,1)=a~e^a$$
$$f(a,2)=a~e^a~(a+1)$$
$$f(a,3)=a~e^a~(a^2+3a+1)$$
$$f(a,4)=a~e^a~(a^3+6a^2+7a+1)$$
$$f(a,5)=a~e^a~(a^4+10a^3+25a^2+15a+1)$$
And so on. The coefficients are Stirling numbers of 2nd kind, and can be found in OEIS: http://oeis.org/A008278.
