[proofplan] We proceed by induction on $n$. The base case $n = 0$ is the [Cauchy Integral Formula](/theorems/345). For the inductive step, we differentiate the integral representation with respect to $z_0$ by passing the derivative under the integral sign. The key computation is justifying the interchange of differentiation and integration via a difference quotient estimate: the integrand $(z - z_0)^{-(n+1)} f(z)$ is a smooth function of $z_0$ when $z$ lies on $\gamma$ and $z_0$ is interior, and the uniform bound on $|z - z_0|$ away from zero (since $z_0 \notin \gamma^*$) provides the domination needed to pass the limit through the integral. Once the derivative formula is established for all $n$, infinite differentiability follows immediately, and the Taylor series representation is obtained by expanding the Cauchy kernel in a geometric series. [/proofplan] [step:Establish the base case $n = 0$ from the Cauchy Integral Formula] The case $n = 0$ asserts \begin{align*} f(z_0) &= \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\, dz, \end{align*} which is exactly the [Cauchy Integral Formula](/theorems/345). The hypotheses of Theorem 345 require $f \in \mathcal{O}(\Omega)$, $\gamma$ a closed contour in $\Omega$ null-homotopic in $\Omega$, and $z_0 \in \Omega \setminus \gamma^*$ with $n(\gamma, z_0) = 1$. All three are given by hypothesis. [guided] The induction starts at $n = 0$, where the formula reads \begin{align*} f(z_0) &= \frac{0!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{0+1}}\, dz = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\, dz. \end{align*} This is the [Cauchy Integral Formula](/theorems/345). We verify its hypotheses: (i) $f \in \mathcal{O}(\Omega)$ is assumed; (ii) $\gamma$ is a closed contour in $\Omega$ that is null-homotopic in $\Omega$, given by hypothesis; (iii) $z_0 \in \Omega \setminus \gamma^*$ with winding number $n(\gamma, z_0) = 1$, also given. Therefore the Cauchy Integral Formula applies and yields the base case. Why is the base case important beyond just "starting the induction"? It provides the integral representation of $f(z_0)$ itself as a contour integral. The entire inductive argument will consist of differentiating this integral representation with respect to $z_0$, extracting each successive derivative from under the integral sign. [/guided] [/step] [step:Differentiate under the integral sign to pass from $n$ to $n+1$] Assume the formula holds for some $n \geq 0$: \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 show $f^{(n)}$ is differentiable at $z_0$ and compute its derivative by forming the difference quotient. Let $h \in \mathbb{C} \setminus \{0\}$ with $|h|$ small enough that $z_0 + h \in \Omega \setminus \gamma^*$. Since $\gamma^*$ is compact and $z_0 \notin \gamma^*$, there exists $\delta > 0$ such that \begin{align*} d := \operatorname{dist}(z_0, \gamma^*) > 0, \end{align*} and we require $|h| < d/2$. The difference quotient is \begin{align*} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} &= \frac{n!}{2\pi i} \oint_\gamma f(z) \cdot \frac{1}{h}\left[\frac{1}{(z - z_0 - h)^{n+1}} - \frac{1}{(z - z_0)^{n+1}}\right] dz. \end{align*} [claim:The difference quotient of the kernel converges uniformly on $\gamma^*$] For all $z \in \gamma^*$, \begin{align*} \frac{1}{h}\left[\frac{1}{(z - z_0 - h)^{n+1}} - \frac{1}{(z - z_0)^{n+1}}\right] \to \frac{n+1}{(z - z_0)^{n+2}} \quad \text{as } h \to 0, \end{align*} and the convergence is uniform on $\gamma^*$. [/claim] [proof] Define $g: \mathbb{C} \setminus \{z_0\} \to \mathbb{C}$ by $g(w) = (w - z_0)^{-(n+1)}$. This function is holomorphic on $\mathbb{C} \setminus \{z_0\}$, and its derivative is \begin{align*} g'(w) &= -(n+1)(w - z_0)^{-(n+2)}. \end{align*} We need to show that $\frac{g(z - h) - g(z)}{-h} \to g'(z)$ uniformly for $z \in \gamma^*$. Fix $z \in \gamma^*$. By the mean value inequality for holomorphic functions (applied to $g'$ on the disk $B(z, d/2)$, which is contained in $\mathbb{C} \setminus \{z_0\}$ since $|z - z_0| \geq d$), for $|h| < d/2$: \begin{align*} \left|\frac{g(z-h) - g(z)}{-h} - g'(z)\right| &= \left|\frac{1}{-h}\int_0^1 [g'(z - th) - g'(z)](-h)\, dt\right| = \left|\int_0^1 [g'(z - th) - g'(z)]\, dt\right|. \end{align*} Here we used the identity $g(z-h) - g(z) = \int_0^1 g'(z - th)(-h)\, dt$, valid because $g'$ is continuous on $\mathbb{C} \setminus \{z_0\}$ and the segment from $z$ to $z - h$ lies in $\mathbb{C} \setminus \{z_0\}$ (since every point on the segment is at distance at least $d - |h| \geq d/2 > 0$ from $z_0$). Now estimate $|g'(z - th) - g'(z)|$. The second derivative of $g$ is $g''(w) = (n+1)(n+2)(w - z_0)^{-(n+3)}$. For $w$ on the segment from $z$ to $z - th$ (where $t \in [0,1]$), we have $|w - z_0| \geq d/2$, so \begin{align*} |g''(w)| &\leq \frac{(n+1)(n+2)}{(d/2)^{n+3}}. \end{align*} By the mean value inequality applied to $g'$: \begin{align*} |g'(z - th) - g'(z)| &\leq |th| \cdot \sup_{|w - z| \leq |h|} |g''(w)| \leq |h| \cdot \frac{(n+1)(n+2)}{(d/2)^{n+3}}. \end{align*} Integrating over $t \in [0,1]$: \begin{align*} \left|\frac{g(z-h) - g(z)}{-h} - g'(z)\right| &\leq |h| \cdot \frac{(n+1)(n+2)}{(d/2)^{n+3}}. \end{align*} The right-hand side is independent of $z \in \gamma^*$ and tends to $0$ as $h \to 0$. Therefore the convergence is uniform on $\gamma^*$. [/proof] By the uniform convergence established in the claim, we may pass the limit $h \to 0$ through the contour integral (the integrand converges uniformly on the compact set $\gamma^*$, and $f$ is continuous on $\gamma^*$, so $f(z)$ times the difference quotient kernel converges uniformly). This gives \begin{align*} \frac{d}{dz_0} f^{(n)}(z_0) &= \frac{n!}{2\pi i} \oint_\gamma f(z) \cdot \frac{n+1}{(z - z_0)^{n+2}}\, dz = \frac{(n+1)!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz. \end{align*} This is the formula for $f^{(n+1)}(z_0)$, completing the inductive step. [guided] The heart of the proof is justifying the passage of the derivative through the integral sign. The inductive hypothesis 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*} We want to differentiate both sides with respect to $z_0$. On the right-hand side, $z_0$ appears only in the integrand $(z - z_0)^{-(n+1)}$. Formally differentiating under the integral sign would give \begin{align*} \frac{d}{dz_0}\left[\frac{1}{(z - z_0)^{n+1}}\right] &= \frac{n+1}{(z - z_0)^{n+2}}, \end{align*} but we must rigorously justify exchanging the limit (of the difference quotient) with the integral. We form the difference quotient with increment $h \in \mathbb{C} \setminus \{0\}$. The key geometric fact is that $z_0$ lies at positive distance $d := \operatorname{dist}(z_0, \gamma^*) > 0$ from the compact image $\gamma^*$ of the contour, since $z_0 \notin \gamma^*$ by hypothesis. We restrict $|h| < d/2$ so that $z_0 + h$ also stays away from $\gamma^*$: for every $z \in \gamma^*$, $|z - z_0 - h| \geq |z - z_0| - |h| \geq d - d/2 = d/2 > 0$. The difference quotient of the kernel is \begin{align*} \frac{1}{h}\left[\frac{1}{(z - z_0 - h)^{n+1}} - \frac{1}{(z - z_0)^{n+1}}\right]. \end{align*} Why does this converge uniformly on $\gamma^*$? Define $g(w) = (w - z_0)^{-(n+1)}$, which is holomorphic on $\mathbb{C} \setminus \{z_0\}$ with derivative $g'(w) = -(n+1)(w - z_0)^{-(n+2)}$. We need to show that $[g(z - h) - g(z)]/(-h) \to g'(z)$ uniformly for $z \in \gamma^*$. Using the integral form of the remainder — $g(z-h) - g(z) = \int_0^1 g'(z - th)(-h)\, dt$ — the error is \begin{align*} \left|\frac{g(z-h) - g(z)}{-h} - g'(z)\right| &= \left|\int_0^1 [g'(z - th) - g'(z)]\, dt\right|. \end{align*} The second derivative satisfies $|g''(w)| \leq (n+1)(n+2)/(d/2)^{n+3}$ for all $w$ with $|w - z_0| \geq d/2$. Since $|z - th - z_0| \geq d - |h| \geq d/2$ for all $t \in [0,1]$ and $z \in \gamma^*$, we get \begin{align*} |g'(z - th) - g'(z)| &\leq |th| \cdot \frac{(n+1)(n+2)}{(d/2)^{n+3}} \leq |h| \cdot \frac{(n+1)(n+2)}{(d/2)^{n+3}}, \end{align*} and after integrating over $t$: \begin{align*} \left|\frac{g(z-h) - g(z)}{-h} - g'(z)\right| &\leq |h| \cdot \frac{(n+1)(n+2)}{(d/2)^{n+3}}. \end{align*} The bound is independent of $z \in \gamma^*$, so the convergence is uniform. Now, since $f$ is continuous on the compact set $\gamma^*$, the product $f(z) \cdot [\text{difference quotient kernel}]$ converges uniformly on $\gamma^*$ to $f(z) \cdot (n+1)(z - z_0)^{-(n+2)}$. Uniform convergence of the integrand on a contour of finite length justifies passing the limit through the integral: \begin{align*} f^{(n+1)}(z_0) &= \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} \\ &= \frac{n!}{2\pi i} \lim_{h \to 0} \oint_\gamma f(z) \cdot \frac{1}{h}\left[\frac{1}{(z - z_0 - h)^{n+1}} - \frac{1}{(z - z_0)^{n+1}}\right] dz \\ &= \frac{n!}{2\pi i} \oint_\gamma f(z) \cdot \frac{n+1}{(z - z_0)^{n+2}}\, dz \\ &= \frac{(n+1)!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz. \end{align*} This completes the inductive step. Note that the argument also shows $f^{(n)}$ is holomorphic (not just differentiable) at $z_0$, since the integral representation defines $f^{(n)}$ as a holomorphic function of $z_0$ on $\Omega \setminus \gamma^*$. [/guided] [/step] [step:Conclude infinite differentiability of $f$] By induction, the derivative formula \begin{align*} f^{(n)}(z_0) &= \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz \end{align*} holds for every $n \in \mathbb{N} \cup \{0\}$. In particular, $f^{(n)}$ exists at every $z_0 \in \Omega$ for every $n \geq 0$: given $z_0 \in \Omega$, choose $r > 0$ with $\overline{B}(z_0, r) \subset \Omega$ and take $\gamma$ to be the circle $|z - z_0| = r$ traversed counterclockwise (which is null-homotopic in $\Omega$ since it bounds a disk contained in $\Omega$, and has winding number $1$ around $z_0$). The inductive step shows each $f^{(n)}$ is itself holomorphic on $\Omega$, so $f \in C^\infty(\Omega)$. [guided] The derivative formula holds for every $n$ by induction. To apply it at a given point $z_0 \in \Omega$, we need a contour $\gamma$ in $\Omega$ enclosing $z_0$ with winding number $1$. Since $\Omega$ is open, there exists $r > 0$ with $\overline{B}(z_0, r) \subset \Omega$. Take $\gamma(t) = z_0 + re^{it}$ for $t \in [0, 2\pi]$. This circle lies in $\Omega$, is null-homotopic in $\Omega$ (it bounds the disk $B(z_0, r) \subset \Omega$), and has $n(\gamma, z_0) = 1$. The formula then expresses $f^{(n)}(z_0)$ as an integral. Since this works for every $n$ and every $z_0$, all derivatives of $f$ exist everywhere in $\Omega$. Moreover, the integral representation shows that each $f^{(n)}$ is itself holomorphic: the map $z_0 \mapsto \oint_\gamma f(z)(z - z_0)^{-(n+1)}\, dz$ is holomorphic in $z_0$ (by the same differentiation-under-the-integral argument applied to $f^{(n)}$ in place of $f$). So each derivative is again holomorphic, which gives $f \in C^\infty(\Omega)$. [/guided] [/step] [step:Expand the Cauchy kernel in a geometric series to obtain the Taylor representation] Fix $z_0 \in \Omega$ and choose $r > 0$ with $\overline{B}(z_0, r) \subset \Omega$. Let $\gamma$ be the circle $|z - z_0| = r$ traversed counterclockwise. For $|z - z_0| = r$ and $|w - z_0| < r$, the ratio $|(w - z_0)/(z - z_0)| < 1$, so the geometric series \begin{align*} \frac{1}{z - w} &= \frac{1}{(z - z_0) - (w - z_0)} = \frac{1}{z - z_0} \cdot \frac{1}{1 - \frac{w - z_0}{z - z_0}} = \sum_{k=0}^{\infty} \frac{(w - z_0)^k}{(z - z_0)^{k+1}} \end{align*} converges absolutely and uniformly in $z$ on $\gamma^*$ for each fixed $w \in B(z_0, r)$. Substituting into the [Cauchy Integral Formula](/theorems/345): \begin{align*} f(w) &= \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - w}\, dz = \frac{1}{2\pi i} \oint_\gamma f(z) \sum_{k=0}^{\infty} \frac{(w - z_0)^k}{(z - z_0)^{k+1}}\, dz. \end{align*} Since $f$ is continuous on the compact set $\gamma^*$ (hence bounded: $|f(z)| \leq M$ for all $z \in \gamma^*$), the series $\sum_{k=0}^{\infty} |f(z)| \cdot |(w - z_0)|^k / |z - z_0|^{k+1} \leq (M/r) \sum_{k=0}^{\infty} (|w - z_0|/r)^k$ converges uniformly in $z \in \gamma^*$. We may therefore interchange the sum and the integral: \begin{align*} f(w) &= \sum_{k=0}^{\infty} (w - z_0)^k \cdot \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{k+1}}\, dz = \sum_{k=0}^{\infty} \frac{f^{(k)}(z_0)}{k!}(w - z_0)^k, \end{align*} where in the last equality we used the derivative formula established above to identify $\frac{1}{2\pi i}\oint_\gamma f(z)(z - z_0)^{-(k+1)}\, dz = f^{(k)}(z_0)/k!$. The series converges for all $w \in B(z_0, r)$, which shows $f$ is represented by its Taylor series in a neighborhood of every point of $\Omega$. [guided] We want to show that $f$ equals its Taylor series locally. The idea is to start from the Cauchy Integral Formula and expand the Cauchy kernel $1/(z - w)$ as a power series in $(w - z_0)$. Fix $z_0 \in \Omega$ and choose $r > 0$ with $\overline{B}(z_0, r) \subset \Omega$. Let $\gamma$ be the circle $|z - z_0| = r$ traversed counterclockwise. For any $w \in B(z_0, r)$, the [Cauchy Integral Formula](/theorems/345) gives \begin{align*} f(w) &= \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - w}\, dz. \end{align*} The hypotheses are satisfied: $f \in \mathcal{O}(\Omega)$, $\gamma$ is null-homotopic in $\Omega$, and $n(\gamma, w) = 1$ since $w \in B(z_0, r)$. Now we expand the kernel. For $z \in \gamma^*$ (so $|z - z_0| = r$) and $w \in B(z_0, r)$ (so $|w - z_0| < r$), the ratio $q := (w - z_0)/(z - z_0)$ satisfies $|q| < 1$. The geometric series formula $1/(1 - q) = \sum_{k=0}^{\infty} q^k$ (valid for $|q| < 1$) gives \begin{align*} \frac{1}{z - w} &= \frac{1}{(z - z_0)(1 - q)} = \sum_{k=0}^{\infty} \frac{(w - z_0)^k}{(z - z_0)^{k+1}}. \end{align*} Can we interchange the sum and the integral? We need uniform convergence in $z$ on $\gamma^*$. The $k$-th term is bounded by \begin{align*} \left|\frac{f(z)(w - z_0)^k}{(z - z_0)^{k+1}}\right| &\leq \frac{M}{r}\left(\frac{|w - z_0|}{r}\right)^k, \end{align*} where $M = \max_{z \in \gamma^*} |f(z)|$ (the maximum exists since $f$ is continuous on the compact set $\gamma^*$). The ratio $|w - z_0|/r < 1$ is fixed, so the geometric series $\sum (M/r)(|w - z_0|/r)^k$ converges. By the Weierstrass $M$-test, the series converges uniformly in $z \in \gamma^*$, and we may interchange sum and integral: \begin{align*} f(w) &= \sum_{k=0}^{\infty} (w - z_0)^k \cdot \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{k+1}}\, dz. \end{align*} The coefficient of $(w - z_0)^k$ is exactly $\frac{1}{2\pi i}\oint_\gamma f(z)(z - z_0)^{-(k+1)}\, dz$, which by the derivative formula (established in the previous steps) equals $f^{(k)}(z_0)/k!$. Therefore \begin{align*} f(w) &= \sum_{k=0}^{\infty} \frac{f^{(k)}(z_0)}{k!}(w - z_0)^k \quad \text{for all } w \in B(z_0, r). \end{align*} This is the Taylor series of $f$ centered at $z_0$, and it converges on the entire disk $B(z_0, r)$. Since $z_0 \in \Omega$ was arbitrary, $f$ is locally represented by its Taylor series everywhere in $\Omega$ — that is, $f$ is analytic on $\Omega$. [/guided] [/step]