TITLE: Derive the EGF of Stirling number of the second kind with the OGF-EGF conversion formula
QUESTION [5 upvotes]: It is well known that, the EGF(Exponential Generating Function) of Stirling number of the second kind is 
$$
\hat G_{k}(x)=\sum_{n=k}^{\infty} S(n, k) \frac{x^{n}}{n !}=\frac{1}{k !}\left(e^{x}-1\right)^{k}
$$
And the OGF(Ordinary Generating Function) of Stirling number of the second kind is 
$$
G_k(x) = \sum_{n\ge k} S(n,k) x^n =
\frac{x^k}{(1-x)(1-2x)\cdots(1-kx)}
$$
The OGF-EGF conversion formula is given here Generating function transformation, that is
$$
\hat F(z)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }F\left(ze^{-\imath \vartheta }\right)e^{e^{\imath \vartheta }}d\vartheta 
$$
where, $\hat F(z)$ means the EGF and $F(z)$ means the OGF.
However, I need to calculate
$$
\int_{-\pi }^{\pi } \frac{e^{e^{i t}} \left(-e^{-i t} z\right)^{-k} \left(e^{-i t} z\right)^k}{\left(1-\frac{e^{i t}}{z}\right){}_k} \, dt, \text{where } (x)_k \text{ the Pochhammer symbol.}
$$
which is  really difficult for me.
So my problem is to calculate the integral above.

REPLY [3 votes]: 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 $(e^z - 1)^k/k!$.
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