Cauchy's integral formula says $ f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-z)^{n+1}}. $
If we let $C$ be the circle of radius $r$, such that $|f(\zeta)|\leq M$ on $C$, then taking $z=a$, one obtains Cauchy's estimate that $ |f^{(n)}(a)|\leq Mn!r^{-n}. $ How is this derived? I see instead $ |f^{(n)}(a)|\leq\frac{n!}{2\pi}\int_C \frac{|f(\zeta)||d\zeta|}{|\zeta-a|^{n+1}}\leq Mn!\int_C\frac{|d\zeta|}{|\zeta-a|^{n+1}} $ but I don't see how this eventually gets to Cauchy's estimate.