Suppose that a function $f$ is analytic in the open disk
$$D_{r}=\{z\in \mathbb{C}:|z|<r\}$$ where $r>0$, and there is a number $M\in\mathbb{R} $ such that $|f(z)| \leq M$ for all $z \in D_{r}$.
Let $\delta \in (0, r)$. Then show that:
$$|f^{n}(z)|\leq \frac{n!M}{\delta^{n}}$$ for all $z\in D_{r−δ}$ where $D_{r−δ} = \{z\in\mathbb{C} : |z| < r − δ\}$.
My try:
The initial disk $D_{r}$ is centered at $z_{0}=0$, the Cauchy formula reduces to
$$|f^{n}(z)|=\frac{n!}{2\pi i}\int_{D_{r}} \frac{f(\zeta)}{(\zeta-z_{0})^{n+1}} d\zeta$$
$$=\frac{n!}{2\pi i}\int_{D_{r}} \frac{f(\zeta)}{\zeta^{n+1}} d\zeta$$
$$\leq \frac{n!}{2\pi }\frac{M}{\delta^{n+1}}2\pi \delta=\frac{n!M}{\delta^{n}}$$
and then extending this theorem, the new disk $D_{r-\delta}$ is still centered at $z_{0}=0$ and the radius is still $R=\delta$. So, you can show that the inequality is true for the new disk by replacing $M=M_{1}$ and show that $M= M_{1}$ equal each other. This doesn't seem like a very complete evaluation of the problem in hand.