Vtyjkyuk

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 $UsubseteqmathbbC$ be an open and path-connected subset, let $z_0in U$, and let $gammasubseteq U$ be some, say, smooth loop such that $z_0notingamma$ and $gammasimeqrmpt.$ in $U$, where by abuse of notation $gamma$ stands for both the curve and its support.
If $f:UtomathbbC$ is a holomorphic function, then:
$$
f(z_0)operatornameind(gamma,z_0) = frac12pi ioint_gammafracf(z)z-z_0mathrmdz
$$
Attempted proof: I know that, being holomorphic, the 1-form
$$
omega:=fracf(z)z-z_0mathrmdz
$$
is $mathrmd$-closed, hence (its integral is) homotopy-invariant.
Since $gammasubseteq Usetminusz_0$ and $[gamma]=0$ in $pi_1(U)$, I believe that $[gamma]$ should induce a well-defined class $[gamma]'$ in $pi_1(Dsetminusz_0)$, 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!