Within in the proof of Cauchy Integral Formula there is this line $$f^{(k)}(z)=\frac{k!}{2 \pi i} \int_\Im \frac{f(\zeta)}{(\zeta - z)^{k+1}} d\zeta \quad (k=1,2,3,...)$$
My goal is to derive this line and my first instinct would be to use induction and doing $f^{(k-1)}(z)$? any guidance would be appreciated. Thanks in advance.
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Differentiate both sides w.r.t. $z$: $$\begin{align*} \frac{d}{dz} f^{(k-1)}(z) &= \frac{d}{dz} \frac{(k-1)!}{2 \pi i} \int_\gamma \frac{f(\zeta)}{(\zeta - z)^k} d\zeta \\ &= \frac{(k-1)!}{2 \pi i} \int_\gamma k (\zeta - z)^{-k-1} f(\zeta) d\zeta \end{align*}$$ The derivative can be passed under the integral sign since the integrand is differentiable (when $\zeta$ is fixed, the integrand is holomorphic everywhere except for $\zeta$, for which $z$ will never equal to) and has bounded derivative.
The proof is by induction on $n$. The Cauchy integral formula gives us for $k=0$, this is: $$f(z)=\frac{1}{2 \pi i} \int_\Im \frac{f(\zeta)}{(\zeta - z)} d\zeta$$
So suppose $k\geq 0$ is an integer and your formula holds for $k$ and for all holomorphic functions in a neighborhood of the region enclosed by the curve $\Im$. We shall show that your formula holds for $k+1$ and for all holomorphic functions in a neighborhood of the region enclosed by the curve $\Im$. In fact, $$ \begin{array}{rcl} f^{(k+1)}(z)&=&(f')^{(k)}(z)=\frac{k!}{2 \pi i} \int_\Im \frac{f'(\zeta)}{(\zeta - z)^{k+1}} d\zeta \\ &=& \frac{(k+1)!}{2 \pi i} \int_\Im \frac{f(\zeta)}{(\zeta - z)^{k+2}} d\zeta \end{array} $$ where the last equality is accomplished by itegratting by parts.