Prove that $\left|f^{(n)} \left(z_0 \right) \right| \leqslant \frac{n! M}{R^n}$ for every $n$. Let $f$ be analytic on a domain $\Omega$ containing the closed disk of radius $R$ centered at $z_0$. Show that if $\left|z-z_0 \right| = R \implies \left| f(z) \right| \leqslant M$ then $\left|f^{(n)} \left(z_0 \right) \right| \leqslant \frac{n! M}{R^n}$ for every $n$.
I tried applying Cauchy's Integral formula derivatives (since $f$ is analytic and $\Omega$ contains the simple closed curve $C_R(z_0)$ =circle of radius $R$ centered at $z_0$): 
Since $f^{(n)}(z_0)= \frac{n!}{2 \pi i} \int_{\Gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz$, we have $$\begin{align*} \left|f^{(n)}(z_0) \right| &= \left|\frac{n!}{2 \pi i} \int_{\Gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz \right| \\&= \left|\frac{n!}{2 \pi i} \right| \int_{C_{R}(z_0)} \frac{\left|f(z) \right|}{\left|(z-z_0) \right|^{n+1}}dz \\& \leqslant \left|\frac{n!}{2 \pi i} \right| \int_{C_{R}(z_0)} \frac{M}{R^{n+1} }dz  \end{align*}.$$
Is this the right way to go about it? I wasn't really sure what to do with the modulus when the integral was involved, but I used facts like $\left|z^n \right|= \left|z \right|^n$ and $\left|uv \right| = \left|u \right| \left|v \right|$. Where do I go from here?

You are done. 
\begin{align}
\left|\frac{n!}{2 \pi i} \right| \int_{C_{R}(z_0)} \frac{M}{R^{n+1} }\,|dz| 
&=\frac{n!}{2\pi}\,\frac{M}{R^{n+1}}\,\int_{C_r(z_0)}\,1\,|dz|
=\frac{n!}{2\pi}\,\frac{M}{R^{n+1}}\,2\pi R
=\frac{n!M}{R^n}.
\end{align}
