Show that $\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{-s(e^{-ik}-1)+ikm}=\frac{s^{m}}{m!}e^{-s}$ Book (Altland and Simons, Ch 10, Eq. 10.41) I am reading used the integral:  
$$\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{-s(e^{-ik}-1)+ikm}=\frac{s^{m}}{m!}e^{-s},\quad m\in\mathbb{Z},\, s\in\mathbb{R}.$$
It says that it is easily shown using Taylor expansion of the exponent and and the identity:
$$\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{ink}=\delta_{n,0},\quad n\in\mathbb{N}.$$
I don't see how to solve this integral using the methods proposed in the book, can somebody fill in the details?
 A: I'll assume that the $-s$ in the integrand's exponent means $s$ instead.
$$\begin{align*}&\phantom{~=}
\int_0^{2\pi}\frac 1 {2\pi}e^{s(e^{-ik}\;-1)+ikm}\,dk\\&=
\int_0^{2\pi}\frac 1 {2\pi}e^{-s+ikm}\left(\sum_{n=0}^\infty \frac 1 {n!}\left(se^{-ik}\right)^n\right)dk\\&=
\sum_{n=0}^\infty \frac {e^{-s}s^n} {n!}\left(\int_0^{2\pi}\frac 1 {2\pi}e^{ikm-ikn}dk\right)\\&=
\sum_{n=0}^\infty \frac {e^{-s}s^n} {n!}\delta_{m,n}\\&=
\frac{e^{-s}s^m}{m!}
\end{align*}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{2\pi}
\,\exp\pars{-s\bracks{\expo{-\ic k} - 1} + \ic km}
{\dd k \over 2\pi} =
\require{cancel}\xcancel{{s^{m} \over m!}\expo{-s}}:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{2\pi}
\,\exp\pars{-s\bracks{\expo{-\ic k} - 1} + \ic km}
{\dd k \over 2\pi}} =
\expo{s}\int_{0}^{2\pi}
{\exp\pars{-s/\expo{\ic k}} \over \pars{\expo{\ic k}}^{-m}}\,
{\dd k \over 2\pi}
\\[5mm] \stackrel{}{=}\,\,\,&\
\expo{s}\oint_{\verts{z} = 1}{\expo{-s/z} \over z^{-m}}\,{\dd k \over 2\pi} =
\expo{s}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,{s^{n} \over n!}\
\oint_{\verts{z} = 1}{1 \over z^{n - m}}
\,{\dd z/\pars{\ic z} \over 2\pi}
\\[5mm] = &\
\expo{s}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,{s^{n} \over n!}\
\underbrace{\oint_{\verts{z} = 1}{1 \over z^{n - m + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{\delta_{nm}}}
\\[5mm] = &\
\expo{s}\bracks{\pars{-1}^{m}\,{s^{m} \over m!}}\bracks{m \in \mathbb{N}_{\geq 0}} =
\bbx{\pars{-1}^{m}\,{s^{m} \over m!}\,\expo{s}
\bracks{m \in \mathbb{N}_{\geq 0}}}
\end{align}
