Let $\gamma:[0,1]\rightarrow G$ be a closed smooth curve in $\mathbb{C}$ such that $n(\gamma;w)=0$ for all $w\in \mathbb{C}-G$ and suppose that $f$ is analytic in $G$. Then $f\circ\gamma=\sigma$ is a closed smooth curve in $\mathbb{C}$. Let $\alpha\in\mathbb{C}$ be such that $\alpha\not\in\{\sigma\}$. We'd like to calculate $n(\sigma;\alpha)$. In Robinson's Functions of One Complex Variable I, he states that $n(\sigma;\alpha)=\frac{1}{2\pi i}\int_\sigma\frac{dw}{w-\alpha}=\frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)-\alpha}dz$.
My question is, how does the second equality follow?
Because\begin{align}\int_\sigma\frac{\mathrm dw}{w-\alpha}&=\int_0^1\frac{\sigma'(t)}{\sigma(t)-\alpha}\,\mathrm dt\\&=\int_0^1\frac{(f\circ\gamma)'(t)}{(f\circ\gamma)(t)-\alpha}\,\mathrm dt\\&=\int_0^1\frac{f'\bigl(\gamma(t)\bigr)f'(t)}{f\bigl(\gamma(t)\bigr)-\alpha}\,\mathrm dt\\&=\int_\gamma\frac{f'(z)}{f(z)-\alpha}\,\mathrm dz.\end{align}