[proofplan]
Since $z_0$ is a pole of order $m$, the function $g(z) := (z - z_0)^m f(z)$ extends to a holomorphic function at $z_0$ with $g(z_0) \neq 0$. We expand $g$ in its Taylor series around $z_0$ using [Taylor's Theorem for Holomorphic Functions](/theorems/348), then divide by $(z - z_0)^m$ to recover the Laurent expansion of $f$. The residue is the coefficient of $(z - z_0)^{-1}$, which corresponds to the $(m-1)$-th Taylor coefficient of $g$, yielding the formula.
[/proofplan]
[step:Express $f$ via the holomorphic function $g(z) = (z - z_0)^m f(z)$]
Since $z_0$ is a pole of order $m$ of $f$, by definition the [Laurent Series Expansion](/theorems/350) of $f$ around $z_0$ takes the form
\begin{align*}
f(z) = \sum_{n=-m}^{\infty} a_n (z - z_0)^n
\end{align*}
with $a_{-m} \neq 0$, valid in some punctured disc $0 < |z - z_0| < R$. Define
\begin{align*}
g: B(z_0, R) &\to \mathbb{C} \\
z &\mapsto (z - z_0)^m f(z) \quad \text{for } z \neq z_0, \qquad g(z_0) := a_{-m}.
\end{align*}
Then for $z \neq z_0$:
\begin{align*}
g(z) = (z - z_0)^m \sum_{n=-m}^{\infty} a_n (z - z_0)^n = \sum_{n=-m}^{\infty} a_n (z - z_0)^{n+m} = \sum_{k=0}^{\infty} a_{k-m} (z - z_0)^k,
\end{align*}
where we substituted $k = n + m$. This is a power series converging on $B(z_0, R)$, so $g$ is holomorphic on $B(z_0, R)$ by [Holomorphicity of Power Series](/theorems/335), with $g(z_0) = a_{-m} \neq 0$.
[/step]
[step:Expand $g$ in its Taylor series and identify the residue as the $(m-1)$-th coefficient]
By [Taylor's Theorem for Holomorphic Functions](/theorems/348), since $g$ is holomorphic on $B(z_0, R)$, it has the Taylor expansion
\begin{align*}
g(z) = \sum_{k=0}^{\infty} \frac{g^{(k)}(z_0)}{k!} (z - z_0)^k.
\end{align*}
By uniqueness of power series expansions, we identify $a_{k-m} = g^{(k)}(z_0)/k!$ for each $k \ge 0$. The residue of $f$ at $z_0$ is
\begin{align*}
\operatorname{Res}(f, z_0) = a_{-1}.
\end{align*}
Setting $k = m - 1$:
\begin{align*}
a_{-1} = a_{(m-1)-m} = \frac{g^{(m-1)}(z_0)}{(m-1)!}.
\end{align*}
[guided]
By [Taylor's Theorem for Holomorphic Functions](/theorems/348), since $g$ is holomorphic on $B(z_0, R)$ (verified in the previous step), it admits the Taylor expansion
\begin{align*}
g(z) = \sum_{k=0}^{\infty} \frac{g^{(k)}(z_0)}{k!} (z - z_0)^k.
\end{align*}
But we also established that $g(z) = \sum_{k=0}^{\infty} a_{k-m} (z - z_0)^k$. Since the power series expansion of a holomorphic function about a point is unique (by [Taylor's Theorem for Holomorphic Functions](/theorems/348)), we may equate coefficients:
\begin{align*}
a_{k-m} = \frac{g^{(k)}(z_0)}{k!} \qquad \text{for all } k \ge 0.
\end{align*}
The residue of $f$ at $z_0$ is, by definition, the coefficient of $(z - z_0)^{-1}$ in the Laurent expansion, namely $a_{-1}$. Which value of $k$ gives $a_{-1}$? We need $k - m = -1$, so $k = m - 1$. Substituting:
\begin{align*}
\operatorname{Res}(f, z_0) = a_{-1} = \frac{g^{(m-1)}(z_0)}{(m-1)!}.
\end{align*}
[/guided]
[/step]
[step:Express the Taylor coefficient as the stated limit]
Since $g$ is holomorphic at $z_0$, the derivative $g^{(m-1)}(z_0)$ is given by the ordinary limit definition:
\begin{align*}
g^{(m-1)}(z_0) = \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} g(z) = \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right],
\end{align*}
where the second equality uses the definition $g(z) = (z - z_0)^m f(z)$ for $z \neq z_0$, and the limit accounts for the fact that $f$ is only defined on the punctured disc. Substituting into the result of the previous step:
\begin{align*}
\operatorname{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right].
\end{align*}
[guided]
We have established that $\operatorname{Res}(f, z_0) = g^{(m-1)}(z_0)/(m-1)!$. It remains to express $g^{(m-1)}(z_0)$ in the form stated in the theorem. Since $g$ is holomorphic on $B(z_0, R)$, the function $g^{(m-1)}: B(z_0, R) \to \mathbb{C}$ is continuous (holomorphic functions are infinitely differentiable by [Holomorphic Implies Infinitely Differentiable](/theorems/3353)), so
\begin{align*}
g^{(m-1)}(z_0) = \lim_{z \to z_0} g^{(m-1)}(z).
\end{align*}
For $z \neq z_0$, we have $g(z) = (z - z_0)^m f(z)$, so
\begin{align*}
g^{(m-1)}(z) = \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right] \qquad \text{for } z \neq z_0.
\end{align*}
Why do we write $\lim_{z \to z_0}$ rather than simply evaluating at $z_0$? Because $f$ is not defined at $z_0$ (it has a pole there), so the expression $(z - z_0)^m f(z)$ only makes sense for $z \neq z_0$. However, $g$ extends holomorphically across $z_0$, and the limit of the $(m-1)$-th derivative exists and equals $g^{(m-1)}(z_0)$. Combining:
\begin{align*}
\operatorname{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right].
\end{align*}
[/guided]
[/step]
