An excerpt from my course notes in introductory complex analysis.
Given a curve $\Gamma$ in the complex plane and a continuous function $\varphi : \Gamma \to \mathbb{C}$, define its Cauchy integral to be $$ f_\varphi : \mathbb{C} \setminus \Gamma \to \mathbb{C} : z \mapsto \frac{1}{2\pi i}\int_\Gamma \frac{\varphi(\zeta)}{\zeta - z}d\zeta . $$ My course notes claim that $f_\varphi$ is then smooth and that its $n$-th derivative is given by $$ \frac{d^n}{dz^n} f_\varphi (z) = \frac{n!}{2\pi i}\int_\Gamma \frac{\varphi(\zeta)}{(\zeta - z)^{n+1}}d\zeta . $$ In the proof, it is shown that the above formula holds for $n = 1$, namely that $$ \frac{d}{dz} f_\varphi (z) = \frac{1}{2\pi i}\int_\Gamma \frac{\varphi(\zeta)}{(\zeta - z)^2}d\zeta $$ and the proof concludes by saying that the general case follows easily by induction on $n$, without elaborating on this.
I tend to write out proofs whenever teachers write '... follows easily' or something similar. Most of the time, the result indeed does follow easily, but this time I'm stuck. How does the general case follow from the case $n = 1$?
Just to clarify: in my course notes, the above result is derived before the Cauchy integral theorem and formula, so neither of those should be used in the proof.
For $n \geqslant 1$, consider the function
$$F_n(\zeta,z) = \frac{\varphi(\zeta)}{(\zeta - z)^n}$$
defined on $\Gamma \times (\mathbb{C}\setminus \Gamma)$. This function is continuous, and it is holomorphic in $z$, in particular it is infinitely often real differentiable with respect to $x = \operatorname{Re} z$ and $y = \operatorname{Im} z$. At the beginning of the course it may not yet be known that holomorphic functions are infinitely often differentiable, but for the instance used here, $z \mapsto \frac{a}{(b-z)^n}$, this is easily checked.
For every $z \in \mathbb{C}\setminus \Gamma$, there is an $r > 0$ such that $D_{2r}(z) = \{ w\in \mathbb{C} : \lvert w-z\rvert < 2r\}$ does not intersect $\Gamma$, and thus the partial derivatives of $F_n$ with respect to $x$ and $y$ are uniformly bounded on $\Gamma \times \overline{D_r(z)}$, hence differentiation under the integral is legitimate, and
$$\frac{\partial}{\partial x} \int_{\Gamma} F_n(\zeta,z)\,d\zeta = \int_{\Gamma} \frac{\partial F_n}{\partial x}(\zeta,z)\,d\zeta,\quad \frac{\partial}{\partial y} \int_{\Gamma} F_n(\zeta,z)\,d\zeta = \int_{\Gamma} \frac{\partial F_n}{\partial y}(\zeta,z)\,d\zeta.$$
For the Wirtinger derivatives $\frac{\partial}{\partial z} = \frac{1}{2}\bigl(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\bigr)$ and $\frac{\partial}{\partial \overline{z}} = \frac{1}{2}\bigl(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\bigr)$ we thus obtain
$$\frac{\partial}{\partial \overline{z}} \int_{\Gamma} F_n(\zeta,z)\,d\zeta = \int_{\Gamma} \frac{\partial F_n}{\partial \overline{z}} (\zeta,z)\,d\zeta = \int_{\Gamma} 0\,d\zeta = 0$$
and
$$\frac{\partial}{\partial z}\int_{\Gamma} F_n(\zeta,z)\,d\zeta = \int_{\Gamma} \frac{\partial F_n}{\partial z}(\zeta,z)\,d\zeta = \int_{\Gamma} n\frac{\varphi(\zeta)}{(\zeta - z)^{n+1}}\,d\zeta,$$
so $I_n(z) = \int_{\Gamma} F_n(\zeta,z)\,d\zeta$ is holomorphic on $\mathbb{C}\setminus \Gamma$ with $\frac{d}{dz} I_n(z) = n\cdot I_{n+1}(z)$, which shows that $f_{\varphi} = \frac{1}{2\pi i} I_1$ is infinitely often complex differentiable and the given formula
$$\frac{d^n}{dz^n} f_{\varphi}(z) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{\varphi(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta$$
holds.