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.
Best Answer
The most conceptual explanation uses composition of species.
Recall that your Stirling number of the first kind counts collections of $k$ cycles that together contain $n$ points.
Now, you first determine the exponential generating function for cycles:
There are $(n-1)!$ cycles of length $n$, for example, because you regard the list of elements coming after 1 in the cycle.
So, you get $Cycles(z)=\sum_n \frac{(n-1)!}{n!}z^n= \sum \frac{z^n}{n}= \log(\frac 1 {1-z})$.
Then, you determine the exponential generating function for sets of size $k$.
There is only one possibility to do so and it has size $k$:
$Sets_k(z)=\frac{z^k}{k!}$
And, now the miracle occurs:
You want to count $k$-sets of cycles and you find the exponential generating function by calculating $Sets_k$ of $Cycles$ with composition of functions.
So, you get $Sets_k(Cycles(z))= \frac{(\log(\frac 1 {1-z}))^k}{k!}$.
You can read more on this on the corresponding Wikipedia page or you can read much more on this in the book Combinatorial Species and Tree-Like Structures by F. Bergeron, Gilbert Labelle, Pierre LeRoux.