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  "body": "\n## Statement\n\nLet $n$ be a natural number, and let $f$ and $g$ be $n$ times differentiable real functions on an open interval $I$.\nThen $f g$ is $n$ times differentiable on $I$, and\n$$(f g)^{(n)} = \\sum_{k=0}^{n} \\binom{n}{k} f^{(k)} g^{(n-k)}.$$\n\n## Remarks\n\nThe proof is an induction on $n$ with the [[product rule|0V9A43H7WV]], like the proof of the binomial theorem.\nFor $n = 1$ it is the product rule. For $n = 2$ it is $(f g)'' = f'' g + 2 f' g' + f g''$.\nMathlib states it as `iteratedDeriv_mul`.\n\n## Explanations\n\n### L1 human\n\nThe rule for the $n$-th derivative of a product looks like the binomial formula for $(a + b)^n$.\nEach term distributes $n$ differentiations between the two factors.\n",
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