I believe there should be a simple topological proof of the Nullhomotopical Cauchy Integral Formula based only on the Cauchy Integral Formula over a Circle, but I can't quite finish the argument and would appreciate some help. (Hopefully I am on the right track.)
Nullhomotopical Cauchy Integral Formula: Let $U\subseteq\mathbb{C}$ be an open and path-connected subset, let $z_0\in U$, and let $\gamma\subseteq U$ be some, say, smooth loop such that $z_0\notin\gamma$ and $\gamma\simeq\rm{pt.}$ in $U$, where by abuse of notation $\gamma$ stands for both the curve and its support. If $f:U\to\mathbb{C}$ is a holomorphic function, then: $$ f(z_0)\operatorname{ind}(\gamma,z_0) = \frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-z_0}\mathrm{d}z $$ Attempted proof: I know that, being holomorphic, the 1-form $$ \omega:=\frac{f(z)}{z-z_0}\mathrm{d}z $$ is $\mathrm{d}$-closed, hence (its integral is) homotopy-invariant. Since $\gamma\subseteq U\setminus\{z_0\}$ and $[\gamma]=0$ in $\pi_1(U)$, I believe that $[\gamma]$ should induce a well-defined class $[\gamma]'$ in $\pi_1(D\setminus\{z_0\})$, where $D$ is a sufficiently small open disk around $z_0$ contained in $U$, but I don't quite see how to finish this line of reasoning rigorously, assuming it actually makes sense. Any help would be appreciated!