Closed form expression for the series $\sum_{k=0}^\infty \frac{k^j}{k!}$ in terms of $j$ For my research I need to compute "approximate" moment generating function (we approximate exponential with the first $D$ elements of its Taylor series) of the number of fixed points of a permutation which involves the following sum:
\begin{equation}\sum_{k=0}^\infty \frac{k^j}{k!} \end{equation}
in terms of $j$. I am pretty sure this grows exponential in $j$ but I am interested in finding the exact exponent. With some elementary calculations I get the following recurrence relation
\begin{equation}\sum_{k=0}^\infty \frac{k^j}{k!}  = \sum_{m=0}^{j-1}{j-1\choose m} \sum_{k=0}^\infty \frac{k^{j-1-m}}{k!}  \end{equation}
but I couldn't proceed further.
 A: This will be
$$
\lim_{x\to1}\left(x\frac d{dx}\right)^j e^x=\lim_{x\to1}e^x\sum_{k=1}^j{j\brace k}x^k= 
B_je,
$$
where ${j\brace k}$ are the Stirling numbers of the second kind and $B_j=\sum_{k=1}^j{j\brace k}$ are the Bell (or exponential) numbers.
A: Note that we have
$$\sum_{k=0}^\infty \frac{k^j}{k!}=\left(\left.\left(x\frac{d}{dx}\right)^j\{e^x\}\right)\right|_{x=1}$$

LEMMA:
In general, if $f(x)$ is $n$ times differentiable, then we have
$$\left(x\frac{d}{dx}\right)^j\{f(x)\}=\sum_{k=1}^j  \begin{Bmatrix} j \\ k \end{Bmatrix}x^k\frac{d^k\,f(x)}{dx^k}$$
where $\begin{Bmatrix} j \\ k \end{Bmatrix}=\frac1{k!}\sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell}\,\ell^j$ are the Stirling Numbers of the Second Kind.

PROOF:
We assume that we can write $\left(x\frac{d}{dx}\right)^j\{f(x)\}$ as 
$$\left(x\frac{d}{dx}\right)^j\{f(x)\}=\sum_{k=1}^j s(j,k)x^k\frac{d^kf(x)}{dx^k}$$
for coefficients $s(j,k)$, where $s(j,1)=1$.  As an inductive base case, take $j=1$.  Then, we have
$$\begin{align}
\left(x\frac{d}{dx}\right)^{j+1}\{f(x)\}&=\left(x\frac{d}{dx}\right)\sum_{k=1}^j s(j,k)x^k\frac{d^kf(x)}{dx^k}\\\\
&=\sum_{k=1}^j s(j,k)\left(x^{k+1}\frac{d^{k+1}f(x)}{dx^{k+1}}+kx^k \frac{d^{k}f(x)}{dx^{k}}\right)\\\\
&=\sum_{k=1}^{j+1} \left(s(j,k-1)+k s(j,k)\right)x^{k}\frac{d^{k}f(x)}{dx^{k}}\\\\
&=\sum_{k=1}^{j+1} s(j+1,k)x^{k}\frac{d^{k}f(x)}{dx^{k}}
\end{align}$$
where $s(j,k)$ satisfies the relationship
$$s(j+1,k)=s(j,k-1)+ks(j,k)$$
along with the conditions $s(j,1)=s(j,j)=1$ and $s(j,k)=0$ for $k>j$.  Noting that  $S(j,k)$ and $\begin{Bmatrix} j \\ k \end{Bmatrix}$ satisfy the same recurrence relationship and initial conditions, the proof is complete.  

If $f(x)=e^x$ then 
$$\begin{align}
\left.\left(\left(x\frac{d}{dx}\right)^j\{e^x\}\right)\right|_{x=1}&=e\sum_{k=1}^j \begin{Bmatrix}j\\k\end{Bmatrix}\\\\
&=eB_j
\end{align}$$
where $B_j=\sum_{k=1}^j \begin{Bmatrix}j\\k\end{Bmatrix}$ are the Bell Numbers. 

Alternatively, we could use the Faà_di_Bruno Formula to calculate the term $\left(x\frac{d}{dx}\right)^j \{e^x\}$.  To do this, we first make the substitution $x=e^y$.  
Then, using the Bell Polynomials, $B_{n,k}(x_1,x_2,\dots,x_{n-k+1})$, we have 
$$\begin{align}
\left(\left.\left(x\frac{d}{dx}\right)^j\{e^x\}\right)\right|_{x=1}&=\left(\left.\left(\frac{d}{dy}\right)^j\{e^{e^y}\}\right)\right|_{y=0}\\\\
&=\sum_{k=1}^j \left(\left.\left(e^{e^y}\right)^{(k)}B_{j,k}((e^y)^{(1)},(e^y)^{(2)},\dots,(e^y)^{(j-k+1)})\right)\right|_{y=0}\\\\
&=e\sum_{k=1}^j B_{j,k}(1,1\dots, 1)\tag1\\\\
&=B_j e\tag2
\end{align}$$
In going from $(1)$ to $(2)$ we used the relationship between the Bell Numbers, $B_n$, and the sum over the Stirling Numbers of the second kind: 
$$B_n=\sum_{k=1}^n B_{n,k}(1,1,\dots,1)=\sum_{k=1}^n {n \brace k}$$

As a side note, it is interesting to observe that 
$$\sum_{j=0}^\infty\frac1{j!} \left(\sum_{k=0}^\infty\frac{k^j}{k!}\right)=e^e$$
A: @user has already answered the question, and @MikeEarnest has noted how little we know about Bell numbers, but I'll chime in with an explanation of where these numbers come from.
In how many ways can we partition a size-$n+1$ set? Call the answer $B_n$. Fix one element $a$ and count the number of solutions in which all but $k$ of the remaining $n$ objects are grouped with $a$; we choose those $k$ in one of $\binom{n}{k}$ ways, so $B_{n+1}=\sum_{k=0}^n\binom{n}{k}B_k$.
It's not hard to show this implies the exponential generating function $B(x):=\sum_n\frac{B_nx^n}{n!}$ satisfies $B(0)=1,\,B^\prime(x)=B(x)\exp x$, so $B(x)=\exp(\exp x-1)$. If we write $B^{(n)}(x)=p_n(\exp x)B(x)$ the polynomials satisfy $p_0=1,\,p_{n+1}(x)=x(p_n(x)+p_n^\prime(x))$, the same recursion relation as the $p_n$ we can define from $\left(x\frac{d}{dx}\right)^n\exp x=p_n(x)\exp x$. This implies these definitions are equivalent, and $p_n(1)=\frac{B^{(n)}(0)}{B(0)}=B_n$, as already stated.
