If $f$ has pole of order $m$, then $\text{res}\left(f,z_0\right)=\lim_{z\to z_0}\frac{1}{(m-1)!}\left\{(z-z_0)^mf(z)\right\}^{(m-1)}$ Statement: Let $$f(z):=\sum_{k=-\infty}^\infty a_kz^k$$ have a pole of order $m$ at $z_0$. Then $$\text{res}\left(f,z_0\right)=\lim_{z\to z_0}\frac{1}{(m-1)!}\left\{(z-z_0)^mf(z)\right\}^{(m-1)}$$

Since $f$ has a pole of order $m$ $$(*)\;\;\;f(z)=\sum_{k=0}^\infty a_kz_k+\sum_{k=1}^m a_{-k}z^{-k}$$ By the Residue theorem we've got $$\text{res}\left(f,z_0\right)\text{ind}_\gamma\text{ }z_0=\frac{1}{2\pi i}\int_\gamma f(z)\;dz$$ where $\gamma$ is a nullhomologous path such that $z_0\notin\gamma^*$. Now comes the point where I'm a little bit confused: Wouldn't Cauchy's Integral theorem yield $\int_\gamma f(z)\;dz=0$? [No, it would not - as stated in the comment of Squirtle and the answer of Daniel Fischer].
However, how to proceed? I'm unsure whether $(*)$ does help here or not. Another fact that follows from the pole of order $m$ at $z_0$ is the existence of a holomorphic function $g$ with $$f(z)=\frac{1}{(z-z_0)^m}g(z)$$ But I still got problems to obtain the statement.

Notes:


*

*$\text{ind}_\gamma\text{ }z_0$ denotes the Winding number

*$\gamma^*$ denotes the trace

 A: 
where $\gamma$ is a nullhomologous path such that $z_0 \notin \gamma^\ast$

The point is: "nullhomologous with respect to which domain"?
Cauchy's integral theorem states that if $h\colon \Omega \to \mathbb{C}$ is holomorphic, then
$$\int_\kappa h(z)\,dz = 0$$
for every closed path of integration (more generally, every cycle) $\kappa$ that is nullhomologous in $\Omega$.
Here, you have $f$ meromorphic on $U$, with a pole in $z_0\in U$ (and, for simplicity, no other pole), and the residue theorem says that
$$\frac{1}{2\pi i} \int_\gamma f(z)\,dz = \operatorname{ind}_\gamma z_0\cdot \operatorname{res}(f,z_0)$$
for every closed path of integration (cycle) $\gamma$ that is nullhomologous in $U$ and does not pass through $z_0$, while Cauchy's integral theorem says
$$\frac{1}{2\pi i} \int_\gamma f(z)\,dz = 0$$
for every closed path of integration (cycle) $\gamma$ that is nullhomologous in $U\setminus \{z_0\}$.
If $\gamma$ is nullhomologous in $U\setminus \{z_0\}$, then it is also nullhomologous in $U$, but additionally $\operatorname{ind}_\gamma z_0 = 0$. So in that case, the residue theorem says exactly the same thing as the integral theorem.
If $\operatorname{ind}_\gamma z_0 \neq 0$, then $\gamma$ can be nullhomologous in $U$, but it is not nullhomologous in $U\setminus \{z_0\}$, so then Cauchy's integral theorem says nothing about the value of the integral.
Now, concerning the formulae
$$\operatorname{res}(f,z_0) = \frac{1}{(m-1)!} \lim_{z\to z_0} \left\{ (z-z_0)^m f(z)\right\}^{(m-1)}$$
and $(\ast)$:
The residue of $f$ in $z_0$ is the coefficient $a_{-1}$ in the Laurent expansion
$$f(z) = \sum_{n=-\infty}^\infty a_n (z-z_0)^n.\tag{0}$$
From the definition of the residue as an integral, we obtain that by expanding the integrand into its Laurent series, and interchanging summation and integration (which is legitimate since the Laurent series converges uniformly on all small enough circles centred at $z_0$). All terms except the $a_{-1}(z-z_0)^{-1}$ terms have a primitive on $\mathbb{C}\setminus\{z_0\}$, and hence vanishing integral.
If $f$ has a pole of order $m$ in $z_0$, then all coefficients $a_n$ for $n < -m$ vanish, and $(0)$ becomes, writing $(\ast)$ as a single series,
$$f(z) = \sum_{n=-m}^\infty a_n (z-z_0)^n.\tag{1}$$
From that, we obtain the Taylor expansion of $g(z) = (z-z_0)^m\cdot f(z)$ about $z_0$ by a simple multiplication:
$$(z-z_0)^mf(z) = \sum_{n=-m}^\infty a_n(z-z_0)^{n+m} = \sum_{k=0}^\infty a_{k-m}(z-z_0)^k = \sum_{k=0}^\infty b_k (z-z_0)^k.\tag{2}$$
By the shifting of the indices, we have
$$\operatorname{res}(f,z_0) = a_{-1} = a_{(m-1)-m} = b_{m-1},$$
so the residue of $f$ in $z_0$ is the $(m-1)^{\text{st}}$ coefficient of the Taylor expansion of $g$, which is
$$b_{m-1} = \frac{g^{(m-1)}(z_0)}{(m-1)!}.$$
Since $g$ is holomorphic, all its derivatives are continuous, and so
$$g^{(m-1)}(z_0) = \lim_{z\to z_0} g^{(m-1)}(z) = \lim_{z\to z_0} \left(\frac{d}{dz}\right)^{m-1}\left((z-z_0)^mf(z)\right)$$
and we have the limit formula.
