My text gives a much more complicated proof of this result, which makes me wonder if the argument I have in my head for this has something wrong with it. Does this work, or have I made a bad assumption somewhere along the line?
Let $U \subseteq \mathbb{C}$ be open, $\overline{D}(z_0, r) \subset U$, and $f : U \rightarrow \mathbb{C}$ be holomorphic with a zero of order $n$ at $z_0$ and no other zeroes in $U$. Taking a power series expansion at $z_0$, $$f(z) = a_n(z-z_0)^n + o((z-z_0)^{n+1}).$$ Differentiating, $$f'(z) = n a_n(z-z_0)^{n-1} + o((z-z_0)^n),$$ so we have $$\lim_{z \rightarrow z_0} \frac{(z-z_0) f'(z)}{f(z)} = \lim_{z \rightarrow z_0} \frac{n a_n(z-z_0)^n + o((z-z_0)^{n+1})}{a_n(z-z_0)^n + o((z-z_0)^{n+1})} = n.$$ Define the function $g : U \rightarrow \mathbb{C}$ by $$g(z) = \begin{cases} (z-z_0) f'(z)/f(z) & \text{if } z \neq z_0 \\ n & \text{otherwise} \end{cases}$$ By the Cauchy integral formula, $$n = g(z_0) = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{g(z)}{z - z_0} dz = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{(z-z_0)f'(z)}{(z-z_0)f(z)} dz= \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{f'(z)}{f(z)} dz$$ as desired.