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  "body": "\n## Proof\n\nWe differentiate the power series term by term.\n\n1. By definition, the [[exponential function|0RMVS7RXZX]] is the [[power series|0VP56B4EMX]] $\\exp(x) = \\sum_{n=0}^{\\infty} a_n x^n$ with center $0$ and $a_n = \\frac{1}{n!}$.\n2. Its radius of convergence is $R = \\infty$, by [[the exponential series has infinite radius of convergence|0PVJRACB9H]]. So the interval $(0 - R, 0 + R)$ is all of $\\R$.\n3. By the theorem that [[power series are differentiable inside the radius of convergence|002X7EK91H]], $\\exp$ is differentiable on $\\R$. For every real $x$, $\\exp'(x) = \\sum_{n=1}^{\\infty} n a_n x^{n-1}$.\n4. For $n \\ge 1$, the definition of the [[factorial|0WFCGVMRRC]] gives $n! = n \\cdot (n-1)!$. So $n a_n = \\frac{n}{n!} = \\frac{1}{(n-1)!}$.\n5. By steps 3 and 4, and the index change $m = n - 1$, $\\exp'(x) = \\sum_{n=1}^{\\infty} \\frac{x^{n-1}}{(n-1)!} = \\sum_{m=0}^{\\infty} \\frac{x^m}{m!} = \\exp(x)$.\n\n## Idea\n\nDifferentiating $x^n/n!$ gives $x^{n-1}/(n-1)!$, the previous term of the same series. So differentiating the whole series shifts it by one place and returns the same series. The infinite radius of convergence makes term-by-term differentiation valid everywhere.\n",
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