Prove a certain relation holds for a holomorphic function. Suppose $f(z)$ is holomorphic on the disk $|z|&lt;R$ and $|f(z)|\leq M$ for all $z$ with $|z|&lt;R$.

Let $0&lt;r&lt;R$. Prove that for any $z$ with $|z| &lt; r$ it holds that

$|f^{(n)}(z)| \leq \frac{M \cdot n!}{(R-r)^n}$

Could someone provide a hint? Thanks a lot!
Let $\gamma$ be the circle forming the boundary of the closed disk $D=\{w\ |\ |w|\leq R\}$, then since $|z| &lt;r &lt;R \implies z \in \text{Int}D$. Then by Cauchy's Integral Formula, or more appropriately Cauchy's Differentiation Formula, we have

$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz \\ \implies \left|f^{(n)}(z)\right|=\left|\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz\right|=\frac{n!}{2\pi}\cdot\left|\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz\right|\leq \frac{n!}{2\pi}\cdot 2\pi R \cdot \sup_{w \in \gamma}\left|\frac{f(z)}{(w-z)^{n+1}}\right| \\ \implies \left|f^{(n)}(z)\right|\leq n! \cdot R\cdot\frac{M}{(R-r)^{n+1}} \leq \frac{M\cdot n!}{(R-r)^n}$$

The final inequality follows from the fact:

$$|w-z| \geq \left||w|-|z|\right|\geq|R-r|$$
Let $\gamma$ be the circle forming the boundary of the closed disk $D=\{w\ |\ |w|\leq R\}$, then since $|z| &lt;r &lt;R \implies z \in \text{Int}D$. Then by Cauchy's Integral Formula, or more appropriately Cauchy's Differentiation Formula, we have

$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz \\ \implies \left|f^{(n)}(z)\right|=\left|\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz\right|=\frac{n!}{2\pi}\cdot\left|\int_{\gamma}\frac{f(z)}{(w-z)^{n+1}}dz\right|\leq \frac{n!}{2\pi}\cdot 2\pi R \cdot \sup_{w \in \gamma}\left|\frac{f(z)}{(w-z)^{n+1}}\right| \\ \implies \left|f^{(n)}(z)\right|\leq n! \cdot R\cdot\frac{M}{(R-r)^{n+1}} \leq \frac{M\cdot n!}{(R-r)^n}$$

The final inequality follows from the fact:

$$|w-z| \geq \left||w|-|z|\right|\geq|R-r|$$