How is Cauchy's estimate derived?

By Cauchy's integral formula you have given, we have $$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-a)^{n+1}}$$ where $C$ is a circle of radius $r$ centered at $a$. Therefore, $C$ can be parametrized as $\zeta=a+re^{i\theta}$, $0\leq \theta\leq 2\pi$, which implies $$|f^{(n)}(a)|=\left|\frac{n!}{2\pi i}\int_0^{2\pi}\frac{f(a+re^{i\theta})rie^{i\theta}d\theta}{(re^{i\theta})^{n+1}}\right|\leq\frac{n!}{2\pi }\int_0^{2\pi}\left|\frac{f(a+re^{i\theta})rie^{i\theta}}{(re^{i\theta})^{n+1}}\right|d\theta$$ $$=\frac{n!}{2\pi }\int_0^{2\pi}\frac{|f(a+re^{i\theta})|}{r^n}d\theta\leq \frac{n!}{2\pi }\int_0^{2\pi}\frac{M}{r^n}d\theta=\frac{Mn!}{r^n}$$ where the last equality follows from $|e^{i\theta}|=1$ and $|i|=1$, and the last inequality follows from the assumption that $|f|\leq M$ on $C$.