Given a circunference $\Gamma$ with radius $R$ and center $z_0$. Prove that if $f(z)$ is analytic in $\Gamma$ and their interior, then \begin{equation*} |f^{(n)}(z_0)|\leq \frac{Mn!}{R^n}, \end{equation*} where $M$ is the maximum value of $|f(z)|$ in $\Gamma$.
Ok, here my attempt:
Since $f$ is analytic, then by Taylor theorem \begin{equation*} f(z) = \sum _{n = 0}^{\infty}\frac{f^{(n)}(z_0)}{n!} (z-z_0)^n, \end{equation*} taking absolute value we obtain \begin{equation*} |f(z)| = \left|\sum _{n = 0}^{\infty}\frac{f^{(n)}(z_0)}{n!} (z-z_0)^n\right| = \sum _{n = 0}^{\infty}|f^{(n)}(z_0)|\frac{|(z-z_0)^n|}{n!}, \end{equation*} but, by definition, $M\geq |f(z)|$ and expanding the summation \begin{equation*} M\geq |f^0(z_0)|\frac{|z-z_0|^0}{0!}+|f^1(z_0)|\frac{|z-z_0|^1}{1!}+\cdots+|f^n(z_0)|\frac{|z-z_0|^n}{n!}+\cdots, \end{equation*} all terms in the summation are potisives, then if we take only the n-term we can assure that \begin{equation*} M\geq |f^{(n)}(z_0)|\frac{|z-z_0|^n}{n!}. \end{equation*}
Ok here is my problem, if I say $|z-z_0|^n\geq R^n$, I'm finished, but $|z-z_0|\geq R$ is false. So, what did I do wrong?.
Hint By The Cauchy Integral Formula $$\begin{equation*} |f^{(n)}(z_0)|= \left| \frac{n!}{2 \pi i}\int_{\Gamma} \frac{f(z)}{|z-z_0|^{n+1}} dz \right| \leq \frac{n!}{2\pi}\int_{\Gamma} \frac{\left|f(z)\right|}{|z-z_0|^{n+1}} dz \end{equation*}$$
Now use $|f(z)| \leq M$ on $\Gamma$. Also, on $\Gamma$ what is $|z-z_0|$?