How to solve $\sum _{n=0}^{\infty }\:\frac{n^a}{n!}$ for any $a \geq 1$ in general? While doing a problem I encountered this summation
$$\sum _{n=1}^{\infty }\:\frac{n^2}{n!}=2e.$$
I used
$$\sum _{n=0}^{\infty }\:\frac{n}{n!}=e$$ 
and 
$$\sum _{n=0}^{\infty }\:\frac{(n-1)^2}{n!}=e$$
these summations to obtain its value. Is there any more general way to do the problem, therefore a general solution for $\sum _{n=0}^{\infty }\:\frac{n^a}{n!}$ for any $ a \geq 1$?
 A: After some playing around, I saw that it is connected to the Bell numbers. That wikipedia article also includes your sum:
$$\frac{1}{e}\sum_{n=0}^\infty \frac{n^a}{n!} = B_a$$
where $B_a$ is the $a$-th Bell number. This equation is called Dobiński's formula. Take a look at its wikipedia page. The proof is not very easy.
A: Recall that the definition of the exponential function is given by
\begin{equation}
e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}.
\end{equation}
The trick now is to note that if we differentiate and multiply by $x$, we get
\begin{equation}
x\frac{\mathrm{d}}{\mathrm{d}x}e^x=\sum_{n=0}^{\infty}\frac{n}{n!}x^n.
\end{equation}
If we do this again, we get
\begin{equation}
x\frac{\mathrm{d}}{\mathrm{d}x}\left(x\frac{\mathrm{d}}{\mathrm{d}x}e^x\right)=\sum_{n=0}^{\infty}\frac{n^2}{n!}x^n.
\end{equation}
In full generality, define $D=x\frac{\mathrm{d}}{\mathrm{d}x}$ to be this differential operation. Then we have
\begin{equation}
\sum_{n=0}^{\infty}\frac{n^a}{n!}=D^a(e^x)\bigg|_{x=1}.
\end{equation}
Since $\frac{\mathrm{d}}{\mathrm{d}x}e^x$, it is easy to show inductively that
\begin{equation}
D^a(e^x)=P_a(x)e^x
\end{equation}
for some polynomial $P_a(x)$, in which case we can write
\begin{equation}
\sum_{n=0}^{\infty}\frac{n^a}{n!}=P_a(1)e.
\end{equation}
Now to actually find the polynomials $P_a(x)$. We can act with $D$ on $P_a(x)e^x$ to obtain
\begin{equation}
P_{a+1}(x)e^x=D\left(P_a(x)e^x\right)=x\frac{\mathrm{d}P_a}{\mathrm{d}x}\,e^x+xP_a(x)e^x,
\end{equation}
and so the polynomial $P_a(x)$ is recursively defined by
\begin{equation}
P_{a+1}(x)=x\left(P_a(x)+\frac{\mathrm{d}P_a}{\mathrm{d}x}\right),
\end{equation}
with $P_0(x)=1$. The first few of these polynomials are given by
\begin{equation}
\begin{split}
P_0(x)&=1,\\
P_1(x)&=x,\\
P_2(x)&=x^2+x,\\
P_3(x)&=x^3+3x+x,\\
P_4(x)&=x^4+6x^3+7x^2+x,\\
P_5(x)&=x^5+10x^4+25x^3+15x^2+x.
\end{split}
\end{equation}
One can inductively show that the polynomials $P_a(x)$ satisfy (this can be left as an exercise)
\begin{equation}
P_{a+1}(x)=x\sum_{k=0}^{a}\binom{a}{k}P_{k}(x),
\end{equation}
so if we define $B_n=P_n(1)$, then we have $B_0=1$ and
\begin{equation}
B_{a+1}=\sum_{k=0}^{a}\binom{a}{k}B_k,
\end{equation}
which is the recursive definition of the so-called Bell numbers. Thus, we have
\begin{equation}
\sum_{n=0}^{\infty}\frac{n^a}{n!}=B_ae.
\end{equation}
