[proofplan] We proceed by induction on $n$. The base case $n = 1$ is established by differentiating the Cauchy integral formula under the integral sign, justified by uniform convergence of the difference quotient on the compact contour. The inductive step uses the same differentiation-under-the-integral argument, advancing from $f^{(n)}$ to $f^{(n+1)}$. Since the resulting integral representation shows that each $f^{(n)}$ is continuous, we conclude $f \in C^\infty(\Omega)$. [/proofplan] [step:Establish the Cauchy integral representation for $f$ on the disk] Fix $z_0 \in \Omega$ and $r > 0$ with $\overline{B}(z_0, r) \subset \Omega$. Let $C_r$ denote the circle $|w - z_0| = r$, traversed once counterclockwise. By the [Cauchy Integral Formula](/theorems/???), for every $z \in B(z_0, r)$, \begin{align*} f(z) = \frac{1}{2\pi i} \oint_{C_r} \frac{f(w)}{w - z}\, dw. \end{align*} This is the case $n = 0$ of the desired formula. [/step] [step:Differentiate under the integral to obtain the formula for $f'(z)$] Fix $z \in B(z_0, r)$. Choose $\delta > 0$ such that $\overline{B}(z, \delta) \subset B(z_0, r)$. For $h \in \mathbb{C}$ with $0 < |h| < \delta$, form the difference quotient: \begin{align*} \frac{f(z + h) - f(z)}{h} = \frac{1}{2\pi i} \oint_{C_r} f(w) \cdot \frac{1}{h} \left( \frac{1}{w - z - h} - \frac{1}{w - z} \right) dw. \end{align*} The integrand simplifies algebraically: \begin{align*} \frac{1}{h}\left(\frac{1}{w - z - h} - \frac{1}{w - z}\right) = \frac{1}{(w - z - h)(w - z)}. \end{align*} We must show that as $h \to 0$, this converges uniformly in $w \in C_r$ to $\frac{1}{(w - z)^2}$. For $w \in C_r$, we have $|w - z| \ge r - |z - z_0| =: d > 0$ (since $z \in B(z_0, r)$). For $|h| < \delta \le d/2$, we have $|w - z - h| \ge |w - z| - |h| \ge d - d/2 = d/2$, so \begin{align*} \left| \frac{1}{(w - z - h)(w - z)} - \frac{1}{(w - z)^2} \right| &= \left| \frac{h}{(w - z - h)(w - z)^2} \right| \\ &\le \frac{|h|}{(d/2) \cdot d^2} = \frac{2|h|}{d^3}. \end{align*} This bound is independent of $w \in C_r$ and tends to $0$ as $h \to 0$. Since $f$ is continuous on the compact set $C_r$, $|f(w)|$ is bounded on $C_r$, say $|f(w)| \le M$ for all $w \in C_r$. The uniform convergence of the integrand justifies passing the limit under the integral: \begin{align*} f'(z) = \frac{1}{2\pi i} \oint_{C_r} \frac{f(w)}{(w - z)^2}\, dw. \end{align*} [guided] The strategy is to differentiate the Cauchy integral formula by passing the limit of the difference quotient under the integral sign. This requires justification: we need the integrand to converge uniformly on $C_r$. Fix $z \in B(z_0, r)$ and define $d := r - |z - z_0| > 0$, which is the distance from $z$ to the circle $C_r$. Choose $\delta = d/2$. For $|h| < \delta$ and $w \in C_r$: - $|w - z| \ge d$ (distance from $z$ to $C_r$), - $|w - z - h| \ge |w - z| - |h| \ge d - d/2 = d/2$. The difference quotient of the integrand is \begin{align*} \frac{1}{h}\left(\frac{1}{w - z - h} - \frac{1}{w - z}\right) = \frac{1}{(w - z - h)(w - z)}. \end{align*} We compare this to the target $\frac{1}{(w-z)^2}$: \begin{align*} \frac{1}{(w - z - h)(w - z)} - \frac{1}{(w - z)^2} = \frac{h}{(w - z - h)(w - z)^2}. \end{align*} The modulus is bounded by $\frac{|h|}{(d/2) \cdot d^2} = \frac{2|h|}{d^3}$, which is uniform in $w \in C_r$ and tends to $0$ as $h \to 0$. Since $f$ is continuous on the compact circle $C_r$, there exists $M > 0$ with $|f(w)| \le M$ for all $w \in C_r$. The integrand $f(w) \cdot \frac{1}{(w-z-h)(w-z)}$ therefore converges uniformly to $f(w) \cdot \frac{1}{(w-z)^2}$ on $C_r$. This uniform convergence justifies exchanging the limit and the contour integral: \begin{align*} f'(z) = \lim_{h \to 0} \frac{1}{2\pi i} \oint_{C_r} \frac{f(w)}{(w-z-h)(w-z)}\, dw = \frac{1}{2\pi i} \oint_{C_r} \frac{f(w)}{(w-z)^2}\, dw. \end{align*} This establishes the $n = 1$ case. Note that the integral representation shows $f'$ is continuous (even holomorphic) on $B(z_0, r)$, since the integrand depends continuously on $z$. [/guided] [/step] [step:Induct on $n$ to establish the formula for all higher derivatives] We prove by induction that for every $n \ge 0$, $f^{(n)}$ exists on $B(z_0, r)$ and satisfies \begin{align*} f^{(n)}(z) = \frac{n!}{2\pi i} \oint_{C_r} \frac{f(w)}{(w - z)^{n+1}}\, dw. \end{align*} The cases $n = 0$ and $n = 1$ are established above. Suppose the formula holds for some $n \ge 1$. Fix $z \in B(z_0, r)$ and let $d = r - |z - z_0| > 0$. For $0 < |h| < d/2$, form the difference quotient of $f^{(n)}$: \begin{align*} \frac{f^{(n)}(z+h) - f^{(n)}(z)}{h} = \frac{n!}{2\pi i} \oint_{C_r} f(w) \cdot \frac{1}{h}\left(\frac{1}{(w-z-h)^{n+1}} - \frac{1}{(w-z)^{n+1}}\right) dw. \end{align*} [claim:The difference quotient of $(w-z)^{-(n+1)}$ converges uniformly to $(n+1)(w-z)^{-(n+2)}$] For $w \in C_r$ and $|h| < d/2$, the algebraic identity \begin{align*} \frac{1}{h}\left(\frac{1}{(w-z-h)^{n+1}} - \frac{1}{(w-z)^{n+1}}\right) - \frac{n+1}{(w-z)^{n+2}} \end{align*} converges to $0$ uniformly in $w \in C_r$ as $h \to 0$. [/claim] [proof] Write $\zeta = w - z$ so $|\zeta| \ge d$, and set $u = h/\zeta$ so $|u| \le |h|/d < 1/2$. Then \begin{align*} \frac{1}{h}\left(\frac{1}{(\zeta - h)^{n+1}} - \frac{1}{\zeta^{n+1}}\right) = \frac{1}{\zeta^{n+2}} \cdot \frac{(1 - u)^{-(n+1)} - 1}{u}. \end{align*} Define $g: B(0, 1) \to \mathbb{C}$ by $g(u) = (1 - u)^{-(n+1)}$. Then $g(0) = 1$ and $g'(0) = n + 1$, so \begin{align*} \frac{g(u) - g(0)}{u} \to g'(0) = n + 1 \quad \text{as } u \to 0. \end{align*} More precisely, by the mean value inequality for holomorphic functions, for $|u| \le 1/2$: \begin{align*} \left|\frac{g(u) - 1}{u} - (n+1)\right| \le C_n |u|, \end{align*} where $C_n$ depends only on $n$ (from the Taylor remainder of $g$ on $|u| \le 1/2$). Therefore \begin{align*} \left|\frac{1}{h}\left(\frac{1}{(\zeta-h)^{n+1}} - \frac{1}{\zeta^{n+1}}\right) - \frac{n+1}{\zeta^{n+2}}\right| \le \frac{C_n |u|}{|\zeta|^{n+2}} \le \frac{C_n |h|}{d^{n+3}}. \end{align*} This bound is independent of $w \in C_r$ and tends to $0$ as $h \to 0$. [/proof] By the uniform convergence established in the claim, we pass the limit under the integral. Since $|f(w)| \le M$ on $C_r$: \begin{align*} f^{(n+1)}(z) &= \frac{n!}{2\pi i} \oint_{C_r} f(w) \cdot \frac{n+1}{(w - z)^{n+2}}\, dw \\ &= \frac{(n+1)!}{2\pi i} \oint_{C_r} \frac{f(w)}{(w - z)^{n+2}}\, dw. \end{align*} This completes the induction. [/step] [step:Conclude that $f \in C^\infty(\Omega)$] Let $z \in \Omega$ be arbitrary. Since $\Omega$ is open, there exist $z_0$ and $r > 0$ with $z \in B(z_0, r)$ and $\overline{B}(z_0, r) \subset \Omega$. The induction above shows that $f^{(n)}$ exists and is given by a contour integral on $B(z_0, r)$ for every $n \ge 0$. The integral representation \begin{align*} f^{(n)}(z) = \frac{n!}{2\pi i} \oint_{C_r} \frac{f(w)}{(w - z)^{n+1}}\, dw \end{align*} exhibits $f^{(n)}$ as a contour integral of a function that is continuous (in fact holomorphic) in $z$ on $B(z_0, r)$, so $f^{(n)}$ is continuous on $B(z_0, r)$. Since $z \in \Omega$ was arbitrary, $f^{(n)} \in C(\Omega)$ for every $n \ge 0$, hence $f \in C^\infty(\Omega)$. [/step]