I'm trying to find an explicit formula for the aforementioned sum (I am not sure that it is possible).
The first values are as follows, with $S_{k}=\sum_{n=0}^{\infty}\frac{n^k}{n!}$ :
$S_0=e$, $S_1=e$, $S_2=2e$, $S_3=5e$, $S_4=15e$, $S_5=52e$.
I couldn't see any obvious pattern from this, but I was able to find a recurrence relation for $S_k$ , by noticing that
$S_{k+2}=\sum_{n=1}^{\infty}\frac{n^{k}\left ( n-1+1 \right )}{\left ( n-1 \right )!}$ $=S_{k+1}+\sum_{n=1}^{\infty}\frac{\left ( n+1 \right )^{k}}{\left ( n-1 \right )!}$.
By expanding $(n+1)^{k}$, we get $S_{k+2}=S_{k+1}+\sum_{i=0}^{k}\binom{k}{i}S_{i+1}$ .
Problem is, I am unable to solve this recurrence relation.
It kind of reminds me of the relation between Bernoulli's numbers : $\sum _{k=0}^{n}{n+1 \choose {k}}B_{k}=0$. It is possible to find an explicit formula for Bernoulli's numbers, but it seems rather complicated to me (see this thread Explicit formula for Bernoulli numbers by using only the recurrence relation )
Perhaps it is possible to use a similar method to find an explicit formula for $S_k$ ?
EDIT : Thanks for the help. We can take the derivative inside the series since f is a power series with an infinite radius of convergence.
f is the exponential function so $f'=f$ :
$xf'\left ( x \right )=xf\left ( x \right )$,
$x\frac{\mathrm{d}}{\mathrm{d} x}\left ( xf'\left ( x \right ) \right )=\left ( x^{2}+x \right )f\left ( x \right )$,
$x\frac{\mathrm{d}}{\mathrm{d} x}\left (\left ( x^{2}+x \right )f\left ( x \right ) \right )=\left ( x^{3}+3x^{2}+x \right )f\left ( x \right )$
I can't seem to find a general formula for $\left ( x\frac{\mathrm{d} }{\mathrm{d} x} \right )^{k}f\left ( x \right )$ though. I'm pretty sure it involves binomial coeffficients...
If we have one, I see that by plugging $x=1$ we get a formula for $S_k$
