A sum of Stirling numbers of the second kind Find a formula (either exact or asymptotic in $N$) for $S(N)$, where
\begin{equation}
S(N) = \sum_{n=N}^\infty \sum_{k=N}^n \sum_{j=0}^k \binom{k}{j} (-1)^{k-j} (1+j)^n \frac{t^n}{n!}.
\end{equation}
Here $N\geq 1$, $\binom{k}{j}$ is the combinatorial symbol, and $t$ is a formal parameter. 
Note: perhaps it is useful to recall the Stirling numbers of the second kind, 
\begin{equation}
\left\{ \begin{array}{c}
n \\ k 
\end{array} \right\} = \frac{1}{k!} \sum_{j=0}^k \binom{k}{j} (-1)^{k-j} j^n,
\end{equation}
and their generating function 
\begin{equation}
\sum_{n=0}^\infty \left\{ \begin{array}{c}
n \\ k 
\end{array} \right\} \frac{x^n}{n!} = \frac{(e^x-1)^k}{k!}.
\end{equation} 
I appreciate comments regarding special functions, asymptotic techniques, or references that could be useful to solve this problem.
 A: Suppose we seek to evaluate
$$S(N) = \sum_{n=N}^\infty
\sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} 
(1+j)^n \frac{t^n}{n!}.$$
This is
$$S(N) = \sum_{n=N}^\infty  \frac{t^n}{n!}
\sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} 
(1+j)^n.$$
Now introduce
$$(1+j)^n =
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((1+j)z) \; dz.$$
We thus get for the inner sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\sum_{j=0}^k {k\choose j} (-1)^{k-j} 
\exp((1+j)z) \; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}} 
(\exp(z)-1)^k\; dz.$$
Substitute this into the middle sum to get
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}} 
\frac{(\exp(z)-1)^{n+1} - (\exp(z)-1)^N}{\exp(z)-2}
\; dz.$$
Now since  $\exp(z)-1$ starts at $z$  the first term drops  out and we
get
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}} 
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
We thus get for the remaining sum
$$\sum_{n=N}^\infty \frac{t^n}{n!} \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}} 
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
Finally note  that $(\exp(z)-1)^N$ starts  at $z^N$ so we  may lower
the initial value of the remaining summation to zero, getting
$$\sum_{n=0}^\infty t^n \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{n+1}} 
\frac{(\exp(z)-1)^N}{2-\exp(z)}
\; dz.$$
What  we have  here is  an annihilated  coefficient extractor  which
finally yields
$$\frac{\exp(t)}{2-\exp(t)} (\exp(t)-1)^N.$$
Now for  the exponential growth rate  of the coefficients on  $t^n$ we
get the distance to the nearest  singularity which is $\log 2.$ So the
radius of convergence of this sum is $|t|<\log 2.$
Observation. Having reached the end of this computation we see that we didn't need to substitute the variable in the integral, which means we could have used the coefficient extractor notation $[z^n]$ throughout. This does not affect the semantics of the computation.

Remark.  There are several  more examples of the  technique of
annihilated    coefficient   extractors    at    this   MSE    link
I  and at  this MSE
link  II  and  also
here             at             this             MSE             link
III.
