The question is: if $f(z)$ is analytic and $|f(z)|\leq M$ for $|z|\leq r$, find an upper bound for$|f^{(n)}(z)|$ in $|z|\leq\frac{r}{2}$.
My attempt: Since $f(z)$ is analytic where $|z|\leq r$, we know that $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw,$$ and $f(z)$ is bounded by $M$. We know that $\bigg|\int_Cf(z)dz\bigg|\leq\max|f(z)|\cdot\text{(length of C)}$, so $$\bigg|f^{(n)}(z)\bigg|=\bigg|\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|\leq n!\cdot M_n\cdot2\pi r,$$ where $M_n:=\max|\frac{f(w)}{(w-z)^{n+1}}|$, for a fixed $z$.
Is this correct? Just finding an upper bound seems like it shouldn't be too difficult, but I feel that I didn't do it correctly. Also, on a somewhat related note, since $f(z)$ is analytic, and its derivatives are analytic, shouldn't $\int_{|z|=r}f^{(n)}(z)dz=0$?
You are on the right track, but the estimate is not enough.
Fix a $z$ in the Disk of $r$. Consider disk $\gamma$ centered at $z$: $$ |w-z|=\frac{r-|z|}{\rho} $$ where $\rho>1$ so that it is completely inside the Disk of $r$. Then \begin{align} \bigg|f^{(n)}(z)\bigg|&=\bigg|\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg| \\ &=\bigg|\frac{n!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{n+1}}dw\bigg| \\ &\leqslant\frac{n!}{2\pi}\int_{\gamma}\frac{M}{|w-z|^{n+1}}|dw| \\ &=\frac{M n!}{2\pi}\int_{0}^{2\pi}\frac{1}{|w-z|^{n+1}}\frac{r-|z|}{\rho}d\theta \\ &=\frac{M n!}{2\pi}\int_{0}^{2\pi}\frac{\rho^n}{(r-|z|)^{n}}d\theta \\ &=\frac{M n!\rho^n}{(r-|z|)^{n}} \end{align} So in $|z|\leq\frac{r}{2}$, there is $$ \bigg|f^{(n)}(z)\bigg|\leqslant\frac{M n!\rho^n}{(r-r/2)^{n}}=\frac{M n!(2\rho)^n}{r^{n}} $$