[proofplan] We apply the [Cauchy Integral Formula for Derivatives](/theorems/2570) to express $f^{(n)}(z_0)$ as a contour integral over the circle $|z - z_0| = r$. Taking the modulus of both sides and applying the ML-inequality — bounding the integrand pointwise and multiplying by the length of the contour — yields the estimate directly. The only hypothesis to verify is that the closed disk $\overline{B}(z_0, r)$ lies inside the domain of holomorphicity, which is given. [/proofplan] [step:Express $f^{(n)}(z_0)$ as a contour integral via the Cauchy Integral Formula for Derivatives] Let $\gamma$ denote the positively oriented circle $|z - z_0| = r$, parametrized by \begin{align*} \gamma: [0, 2\pi] &\to \mathbb{C} \\ t &\mapsto z_0 + r e^{it}. \end{align*} Since $f \in \mathcal{O}(\Omega)$ and $\overline{B}(z_0, r) \subset \Omega$, the [Cauchy Integral Formula for Derivatives](/theorems/2570) applies: the function $f$ is holomorphic on $\Omega$, the point $z_0$ lies inside the circle $\gamma$, and $\gamma$ together with its interior is contained in $\Omega$. The formula gives \begin{align*} f^{(n)}(z_0) &= \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz. \end{align*} [guided] The [Cauchy Integral Formula for Derivatives](/theorems/2570) states that if $f$ is holomorphic on an open set containing a simple closed contour $\gamma$ and its interior, then for any point $z_0$ inside $\gamma$ and any $n \in \mathbb{N}$, \begin{align*} f^{(n)}(z_0) &= \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz. \end{align*} We need to verify the hypotheses. The contour $\gamma$ is the circle $|z - z_0| = r$, whose interior is the open disk $B(z_0, r)$ and whose closure (contour plus interior) is the closed disk $\overline{B}(z_0, r)$. The hypothesis $\overline{B}(z_0, r) \subset \Omega$ guarantees that $f$ is holomorphic on an open set containing $\gamma$ and its interior. The point $z_0$ is the center of the disk, so it lies inside $\gamma$. All conditions are satisfied. Why start with this formula? The Cauchy Integral Formula converts the local, pointwise quantity $f^{(n)}(z_0)$ — which involves differentiation — into a global, integral quantity over a contour. Integrals are amenable to estimation (bounding the integrand and multiplying by the length of the path), whereas derivatives are not directly bounded by the values of the function. This conversion from differentiation to integration is the mechanism behind all of the Cauchy-type estimates in complex analysis. [/guided] [/step] [step:Bound the modulus by estimating the integrand on $\gamma$ via the ML-inequality] Taking the modulus of both sides of the integral representation: \begin{align*} \left|f^{(n)}(z_0)\right| &= \frac{n!}{2\pi} \left|\oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz\right|. \end{align*} We apply the ML-inequality for contour integrals, which states that $\left|\oint_\gamma g(z)\, dz\right| \leq \sup_{z \in \gamma^*} |g(z)| \cdot \operatorname{length}(\gamma)$ for any continuous $g$ on the image $\gamma^*$ of the contour. We identify $g(z) = f(z)/(z - z_0)^{n+1}$ and estimate each factor on $\gamma^*$. For every $z$ on the circle $|z - z_0| = r$: - $|f(z)| \leq \sup_{|w - z_0| = r} |f(w)| = M_r$, by definition of $M_r$. - $|z - z_0|^{n+1} = r^{n+1}$, since $|z - z_0| = r$ on $\gamma$. Therefore \begin{align*} \sup_{z \in \gamma^*} \left|\frac{f(z)}{(z - z_0)^{n+1}}\right| &\leq \frac{M_r}{r^{n+1}}. \end{align*} The length of the circle $\gamma$ is $\operatorname{length}(\gamma) = 2\pi r$. Combining these bounds: \begin{align*} \left|f^{(n)}(z_0)\right| &\leq \frac{n!}{2\pi} \cdot \frac{M_r}{r^{n+1}} \cdot 2\pi r = \frac{n!\, M_r}{r^n}. \end{align*} This is the desired estimate. [guided] Having expressed $f^{(n)}(z_0)$ as a contour integral, we now estimate its modulus. The standard tool is the **ML-inequality** (sometimes called the estimation lemma): for a continuous function $g$ on the image of a contour $\gamma$, \begin{align*} \left|\oint_\gamma g(z)\, dz\right| &\leq \sup_{z \in \gamma^*} |g(z)| \cdot \operatorname{length}(\gamma). \end{align*} This follows from the triangle inequality applied to the Riemann sums defining the contour integral. The "M" stands for the maximum of $|g|$ and the "L" stands for the length of $\gamma$. We apply this with $g(z) = f(z)/(z - z_0)^{n+1}$. On the circle $|z - z_0| = r$, the denominator has constant modulus: $|z - z_0|^{n+1} = r^{n+1}$. This is the key simplification that makes the circle the natural contour for this estimate — on any other curve, $|z - z_0|$ would vary and we would get a weaker bound. For the numerator, $|f(z)| \leq M_r$ by definition. Therefore the supremum of $|g|$ on $\gamma^*$ is at most $M_r / r^{n+1}$, and the length of the circle is $2\pi r$. Assembling these: \begin{align*} \left|f^{(n)}(z_0)\right| &= \frac{n!}{2\pi} \left|\oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M_r}{r^{n+1}} \cdot 2\pi r. \end{align*} The factors of $2\pi$ cancel, and one power of $r$ from the length cancels against the $r^{n+1}$ in the denominator, leaving \begin{align*} \left|f^{(n)}(z_0)\right| &\leq \frac{n!\, M_r}{r^n}. \end{align*} Note the structure of the bound: the factorial $n!$ comes from the Cauchy integral formula, the supremum $M_r$ controls the size of $f$ on the circle, and the factor $r^{-n}$ captures the decay — the larger the circle on which we can control $f$, the smaller the bound on the derivative. This trade-off between the radius $r$ and the supremum $M_r$ is what makes the Cauchy estimates powerful: for entire functions with global bounds (as in [Liouville's Theorem](/theorems/346)), one can send $r \to \infty$ and force all derivatives to vanish. [/guided] [/step]