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  "body": "\n## Proof\n\nWe differentiate the exponential series term by term.\n\n1. By definition, $\\exp z = \\sum_{n \\ge 0} a_n z^n$ with $a_n = 1/n!$, for each $z \\in \\C$.\n2. The [[radius of convergence|045EWAXY9P]] of this series is $\\sup \\set{r \\ge 0 : (r^n/n!)_{n \\ge 0} \\text{ is bounded}}$. This is the formula for the real series $\\sum x^n/n!$, whose radius is $\\infty$ by [[the exponential series has infinite radius of convergence|0PVJRACB9H]]. So the disc of convergence is $\\C$.\n3. By [[power series are holomorphic in the disc of convergence|0YMP521ZR5]], $\\exp$ is [[complex differentiable|0D1C9Y1EXA]] at each $z \\in \\C$, and\n$$\\exp'(z) = \\sum_{n=1}^{\\infty} n a_n z^{n-1} = \\sum_{n=1}^{\\infty} \\frac{n}{n!} z^{n-1}.$$\n4. For $n \\ge 1$ we have $n / n! = 1/(n-1)!$.\n5. Put $m = n - 1$. The series becomes $\\sum_{m=0}^{\\infty} z^m / m!$, which is $\\exp z$.\n6. So $\\exp'(z) = \\exp z$ for each $z \\in \\C$.\n\n## Idea\n\nDifferentiating $z^n/n!$ gives $z^{n-1}/(n-1)!$, the previous term of the same series. So term-by-term differentiation only shifts the index. The theorem on power series makes term-by-term differentiation legal.\n",
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