yeah this page is just gonna derive the formula
\[f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d\zeta\]where $f$ is holomorphic on some domain $\Omega$ and $D$ is some disc whose closure is completely in $\Omega$ too so we can integrate on the boundary
proof is by induction, base case $n=0$ is true due to the cauchy integral
but assuming truth for $n$, we now try to prove truth for $n+1$ using the straight definition of the derivative:
\[\begin{align*} f^{(n+1)}(z) &= \lim_{h \to 0} \frac{f^{(n)}(z+h)-f^{(n)}(z)}{h} \\ &= \lim_{h \to 0} \frac{1}{h}\left(\frac{n!}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{(\zeta-z-h)^{n+1}} d\zeta - \frac{n!}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d\zeta\right) \\ &= \lim_{h \to 0} \frac{n!}{2 \pi i h} \int_{\partial D} f(\zeta)\left(\frac{1}{(\zeta-z-h)^{n+1}} - \frac{1}{(\zeta-z)^{n+1}}\right) d\zeta \\ &= \lim_{h \to 0} \frac{n!}{2 \pi i} \int_{\partial D} f(\zeta)\left(\frac{1}{h}\left(\frac{1}{(\zeta-z-h)^{n+1}} - \frac{1}{(\zeta-z)^{n+1}}\right)\right) d\zeta \end{align*}\]we're getting somewhat close- now i'm gonna prove this little thing:
\[\lim_{h \to 0} \frac{1}{h}\left(\frac{1}{(\zeta-z-h)^{n+1}} - \frac{1}{(\zeta-z)^{n+1}}\right) =\frac{n+1}{(\zeta-z)^{n+2}}\]lemme first work on the expression without the limit here. there's some bashing here:
\[\begin{align*} & \frac{1}{h}\left(\frac{1}{(\zeta-z-h)^{n+1}} - \frac{1}{(\zeta-z)^{n+1}}\right) \\ ={} & \frac{1}{h} \cdot \left(\frac{1}{\zeta-z-h} - \frac{1}{\zeta-z}\right) \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^{n-k}} \cdot \frac{1}{(\zeta-z-h)^k} \\ ={} & \frac{1}{h} \cdot \frac{h}{(\zeta-z-h)(\zeta-z)} \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^{n-k}} \cdot \frac{1}{(\zeta-z-h)^k} \\ ={} & \frac{1}{(\zeta-z-h)(\zeta-z)} \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^{n-k}} \cdot \frac{1}{(\zeta-z-h)^k} \\ \end{align*}\]but it turns out alright. midway i've used this factorization:
\[a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)\]with $\frac{1}{(\zeta-z-h)^{n+1}}$ as $a$ and $\frac{1}{(\zeta-z)^{n+1}}$ as $b$
after this we can basically just straight plug in $h$ since it doesn't appear in the denominator by itself anymore:
\[\begin{align*} & \lim_{h \to 0} \frac{1}{(\zeta-z-h)(\zeta-z)} \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^{n-k}} \cdot \frac{1}{(\zeta-z-h)^k} \\ ={} & \frac{1}{(\zeta-z-0)(\zeta-z)} \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^{n-k}} \cdot \frac{1}{(\zeta-z-0)^k} \\ ={} & \frac{1}{(\zeta-z)^2} \cdot \sum_{k=0}^n \frac{1}{(\zeta-z)^n} \\ ={} & \frac{n+1}{(\zeta-z)^{n+2}} \end{align*}\]with that out of the way, i think we can finally continue our previous evaluation:
\[\begin{align*} f^{(n+1)}(z) &= \lim_{h \to 0} \frac{n!}{2 \pi i} \int_{\partial D} f(\zeta)\left(\frac{1}{h}\left(\frac{1}{(\zeta-z-h)^{n+1}} - \frac{1}{(\zeta-z)^{n+1}}\right)\right) d\zeta \\ &= \frac{n!}{2 \pi i}\int_{\partial D} f(\zeta) \cdot \frac{n+1}{(\zeta-z)^{n+2}} d\zeta \\ &= \frac{(n+1)!}{2 \pi i}\int_{\partial D} \frac{f(\zeta)}{(\zeta-z)^{n+2}} d\zeta\quad\square \end{align*}\]not sure if a uniform convergence argument is needed here since i think we're doing some limit swapping, but i'm not really sure how to apply it since it's not really a discrete kinda limit here