There are a couple of things going on here.
- You should have $\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^r$ as the final answer. This is because you should have had $\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}e^{kx}$ when you expanded $(e^x-1)^n$.
- The question as you posed it is not asking about the Stirling numbers of the second kind. As Michael Hardy points out in a comment, the Stirling numbers of the second kind count the number of ways to distribute $r$ distinct objects into $n$ indistinguishable boxes. (Often this is phrased as "sets" rather than "indistinguishable boxes," as the former seems clearer.) Your question involves distinct boxes.
Let's expand on the second issue, as the two problems are clearly related. Since there's some notational confusion going on, let's let $T(r,n)$ denote the number of ways to distribute $r$ distinct objects into $n$ distinct boxes with no empty box (i.e., your problem), and we'll let $S(r,n)$ denote the number of ways to distribute $r$ distinct objects into $n$ indistinguishable boxes with no empty box. Then we have the relationship $T(r,n) = n! S(r,n)$. This is because the indistinguishable boxes can be made distinguishable by applying $n$ different labels to them, and there are $n!$ ways to assign $n$ labels to $n$ boxes.
This fits with your computations above. Your (corrected) computations have $$T(r,n) = \sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^r.$$
The formula you cite for the Stirling numbers of the second kind has $$S(r,n) = \frac{1}{n!}\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-k)^r.$$
Since $\binom{n}{n-k} = \binom{n}{k}$, though, by reindexing the sum we get $$S(r,n) = \frac{1}{n!}\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^r,$$
in agreement with the argument above that $T(r,n) = n!S(r,n)$.
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
Let $\zeta = e^{i \vartheta}$, then $$\widehat F(z) = \frac 1 {2 \pi i} \int_{|\zeta| = 1} \frac {e^\zeta F {\left( \frac z \zeta \right)}} \zeta d\zeta = \frac 1 {2 \pi i} \int_{|\zeta| = 1} \frac {e^\zeta} {\zeta \prod_{j = 1}^k \left( \frac \zeta z - j \right)} d\zeta, \quad |z| < \frac 1 k.$$ All the poles are inside the unit circle and are simple. The residues of the integrand are $$\operatorname*{Res}_{\zeta = 0} = \frac {(-1)^k} {k!}, \\ \operatorname*{Res}_{\zeta = l z} = \frac {e^{l z}} {l \prod_{j = 1}^{l - 1} (l - j) \prod_{j = l + 1}^k (l - j)} = \frac {e^{l z}} {l (l - 1)! (-1)^{k - l} (k - l)!} = \\ \frac {(-1)^{k - l}} {k!} \binom k l e^{l z}, \quad 1 \leq l \leq k.$$ The sum of the residues is the binomial expansion of $(e^z - 1)^k/k!$.