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  "body": "\n## Proof\n\nWe apply the estimation lemma to the integral formula for $f^{(n)}(c)$.\n\n1. Since $f$ is [[holomorphic|0DG0Z72CQD]] and the closed disc lies in $U$, the [[formula for the derivatives|0F0H5P5DED]] with $a = c$ gives $f^{(n)}(c) = \\frac{n!}{2 \\pi i} \\oint_{\\abs{z - c} = r} \\frac{f(z)}{(z - c)^{n+1}} \\, dz$.\n2. For $z$ on the circle, $\\abs{z - c} = r$. Since the [[modulus is multiplicative|0HYNZE940T]], $\\abs{\\frac{f(z)}{(z - c)^{n+1}}} = \\frac{\\abs{f(z)}}{r^{n+1}} \\le \\frac{M}{r^{n+1}}$.\n3. The [[circle integral|0T7ABX4CWE]] uses the path $\\gamma(\\theta) = c + r e^{i \\theta}$, $\\theta \\in [0, 2\\pi]$. Since $\\abs{\\gamma'(\\theta)} = \\abs{i r e^{i \\theta}} = r$, its length is $\\int_0^{2\\pi} r \\, d\\theta = 2 \\pi r$.\n4. By the [[estimation lemma|0DJMGPN8K9]], $\\abs{\\oint_{\\abs{z - c} = r} \\frac{f(z)}{(z - c)^{n+1}} \\, dz} \\le \\frac{M}{r^{n+1}} \\cdot 2 \\pi r = \\frac{2 \\pi M}{r^n}$.\n5. By steps 1 and 4, $\\abs{f^{(n)}(c)} \\le \\frac{n!}{2 \\pi} \\cdot \\frac{2 \\pi M}{r^n} = \\frac{n! \\, M}{r^n}$.\n\n## Idea\n\nThe derivative $f^{(n)}(c)$ is an average over the circle of $f(z)/(z - c)^{n+1}$, up to the factor $n!$. On the circle this quantity has size at most $M / r^{n+1}$, and the circle has length $2 \\pi r$.\n",
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