Question 2.5.8 in Mardsen and Hoffman's Basic Complex Analysis:
Let f be entire and let $|f(z)|<M$ for $z$ on the circle $|z|=R$, let $R$ be fixed, prove that
$|f^{(k)}(re^{i\theta})|<\frac{k!M}{(R-r)^k}$, $k=0,1,2,...$ for all $0\leq r < R$.
I was able to do the following:
$$|f^{(k)}(re^{i\theta})|=\frac{k!}{2\pi}\int_{\Gamma}\frac{f(z)}{(z-re^{i\theta})^{k+1}}dz \leq\frac{k!}{2\pi}\int_{\Gamma}\frac{|f(z)|}{(|z|-|re^{i\theta}|)^{k+1}}dz =\\ \frac{k!}{2\pi}\int_{\Gamma}\frac{|f(z)|}{(R-r)^{k+1}}dz\leq\frac{k!}{2\pi }\frac{2\pi RM}{(R-r)^{k+1}} = \frac{k!M}{(R-r)^k}\frac{R}{(R-r)}$$
Using Cauchy's Integral formula and ML-inequality and taking $\Gamma$ to be the circle in quection. I am however stuck with the $\frac{R}{(R-r)}$ term , I suppose I should somehow conclude that it is smaller that 1, but I don't thing this is true. Can someone see any mistake in my solution? Thank You!
Your basic approach is correct, however you have to integrate over a different contour:
Let $\Gamma := \partial B_{R-r}(z_0)$ be the circle with radius $R-r$ around a fixed point $z_0 = r e^{i\theta}$.
Due to the usual Cauchy integral estimate for $f$, one still has that $|f(z)| \leq M$ for all $z\in\Gamma$ (the circle $\Gamma$ is contained inside the ball enclosed by the circle with $|z|=R$).
Then, of course, $|z-z_0|= R-r$ for all $z\in\Gamma$, and so (with the same steps as you did) \begin{align} |f^{(k)}(z_0)| &=\left| \frac{k!}{2\pi i}\int\limits_\Gamma \frac{f(z)}{(z-z_0)^{k+1}}d z\right|\\ &\leq \frac{k!}{2\pi}\int\limits_\Gamma \frac{|f(z)|}{|z-z_0|^{k+1}}d z\\ &= \frac{k!}{2\pi(R-r)^{k+1}}\int\limits_\Gamma |f(z)| dz\\ &\leq \frac{k!}{2\pi(R-r)^{k+1}}\cdot 2\pi (R-r)M\\ &= \frac{M k!}{(R-r)^k}. \end{align}