To solve this problem, we need to prove that if a function $f(z)$ is analytic within a disk $|z - z_0| < R$ and $|f(z)| \leq M$ within this disk, then the inequality

$|f^{(n)}(z_0)| \leq \frac{n! M}{R^n}$

holds for $n = 1, 2, \ldots$.

Here's a step-by-step proof:

1. Using Cauchy's Integral Formula for Derivatives: For a function $f(z)$ analytic inside and on a circle $|z - z_0| = r < R$, the $n$-th derivative of $f(z)$ at $z_0$ is given by

   $f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{|z - z_0| = r} \frac{f(z)}{(z - z_0)^{n+1}} \, dz.$

2. Applying the Maximum Modulus Principle: Since $|f(z)| \leq M$ for $|z - z_0| < R$, this also holds for $|z - z_0| = r$ (by the maximum modulus principle).

3. Bounding the Integral: The magnitude of $f^{(n)}(z_0)$ can be bounded using the modulus of the integral:

   $|f^{(n)}(z_0)| = \left| \frac{n!}{2\pi i} \int_{|z - z_0| = r} \frac{f(z)}{(z - z_0)^{n+1}} \, dz \right| \leq \frac{n!}{2\pi} \int_{|z - z_0| = r} \left| \frac{f(z)}{(z - z_0)^{n+1}} \right| \, |dz|.$

4. Simplifying the Bound: Using $|f(z)| \leq M$ and $|z - z_0| = r$, we have:

   $|f^{(n)}(z_0)| \leq \frac{n!}{2\pi} \int_{|z - z_0| = r} \frac{M}{r^{n+1}} \, |dz|.$

5. Evaluating the Contour Integral: The length of the contour $|z - z_0| = r$ is $2\pi r$:

   $|f^{(n)}(z_0)| \leq \frac{n!}{2\pi} \cdot \frac{M}{r^{n+1}} \cdot 2\pi r = \frac{n! M}{r^n}.$

6. Choosing $r$ and Final Bound: To obtain the desired bound, we select $r$ such that $r = R$:

   $|f^{(n)}(z_0)| \leq \frac{n! M}{R^n}.$

Therefore, we have proven that

$|f^{(n)}(z_0)| \leq \frac{n! M}{R^n},$

for $n = 1, 2, \ldots$.

Would you like any further details or have any questions about this proof?

Here are 5 questions you might want to ask:

1. How does Cauchy's Integral Formula work for higher-order derivatives?
2. Can you explain the Maximum Modulus Principle in more detail?
3. What are the applications of this inequality in complex analysis?
4. How do different values of $R$ and $M$ affect the inequality?
5. Can this result be extended to other types of domains or functions?

Tip: Understanding the geometric interpretation of analytic functions and their derivatives can provide deeper insights into complex analysis.