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  "body": "\n## Proof\n\nWe expand $f$ in a power series around $a$ using the circle $\\abs{z - c} = r$, and read off the derivatives.\n\n1. Put $\\delta = r - \\abs{a - c} \u003e 0$ and $D_a = \\set{w : \\abs{w - a} \u003c \\delta}$. For $w \\in D_a$, $\\abs{w - c} \\le \\abs{w - a} + \\abs{a - c} \u003c r$ by the [[triangle inequality|0FJZMBX0ZZ]].\n2. For $w \\in D_a$, [[Cauchy's integral formula|0DG10TVYPN]] gives $f(w) = \\frac{1}{2 \\pi i} \\oint_{\\abs{z - c} = r} \\frac{f(z)}{z - w} \\, dz$, a [[circle integral|0T7ABX4CWE]].\n3. For $z$ on the circle, $\\abs{z - a} \\ge \\abs{z - c} - \\abs{a - c} = \\delta$. Put $u = (w - a)/(z - a)$. Then $\\abs{u} \\le q$, where $q = \\abs{w - a}/\\delta \u003c 1$.\n4. Since $z - w = (z - a)(1 - u)$, the [[finite geometric sum|05APWSX53S]] gives $\\frac{f(z)}{z - w} = \\sum_{n=0}^{N} \\frac{f(z)}{(z - a)^{n+1}} (w - a)^n + T_N(z)$, with $T_N(z) = \\frac{f(z) \\, u^{N+1}}{(z - a)(1 - u)}$.\n5. The circle is [[compact|07PFYV65MB]], so $\\abs{f} \\le M$ on it for some $M$, by the [[extreme value theorem|0V4DET59NW]]. Since $\\abs{1 - u} \\ge 1 - q$, we get $\\abs{T_N(z)} \\le \\frac{M q^{N+1}}{\\delta (1 - q)}$.\n6. Put $b_n = \\frac{1}{2 \\pi i} \\oint_{\\abs{z - c} = r} \\frac{f(z)}{(z - a)^{n+1}} \\, dz$. Integrating step 4 and using the [[estimation lemma|0DJMGPN8K9]] with length $2 \\pi r$, we get $\\abs{f(w) - \\sum_{n=0}^{N} b_n (w - a)^n} \\le \\frac{r M q^{N+1}}{\\delta (1 - q)}$.\n7. Since [[powers of $q$ tend to zero|0Q7K4AXE53]], $f(w) = \\sum_{n=0}^{\\infty} b_n (w - a)^n$ for every $w \\in D_a$.\n8. For $0 \u003c s \u003c \\delta$ the series converges at $w = a + s$, so $(\\abs{b_n} s^n)$ is bounded. So its [[radius of convergence|045EWAXY9P]] is at least $\\delta$. Let $S$ be its sum on $D_a$.\n9. Since [[power series are holomorphic in the disc of convergence|0YMP521ZR5]], $S'$ is again a power series on $D_a$ with the same radius. By induction on $k$, $S^{(k)}(w) = \\sum_{n \\ge k} n (n-1) \\cdots (n - k + 1) b_n (w - a)^{n-k}$ on $D_a$.\n10. At $w = a$, only the term $n = k$ survives. So $S^{(k)}(a) = k! \\, b_k$.\n11. The derivatives $f^{(k)}$ exist on $U$, since [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]]. We show $f^{(k)} = S^{(k)}$ on $D_a$ by induction. For $k = 0$ this is step 7.\n12. If $f^{(k)} = S^{(k)}$ on the open set $D_a$, then for $w \\in D_a$ the difference quotients of both at $w$ agree for small $h$. So $f^{(k+1)}(w) = S^{(k+1)}(w)$.\n13. By steps 10 and 12, $f^{(n)}(a) = n! \\, b_n = \\frac{n!}{2 \\pi i} \\oint_{\\abs{z - c} = r} \\frac{f(z)}{(z - a)^{n+1}} \\, dz$.\n\n## Idea\n\nThe circle $\\abs{z - c} = r$ stays at distance at least $\\delta$ from $a$. So the Cauchy kernel $1/(z - w)$ can be expanded in powers of $w - a$ for $w$ near $a$. This writes $f$ as a power series around $a$, and the $n$-th coefficient of a power series is the $n$-th derivative divided by $n!$.\n",
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