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  "body": "\n## Proof\n\nWe compute $e^{2 \\pi i}$ and use the product rule for exponentials.\n\n1. By [[Euler's formula|04R3CY41VJ]] with $z = 2\\pi$, we get $e^{2 \\pi i} = \\cos 2\\pi + i \\sin 2\\pi$. Here $e^w$ is the [[complex exponential|054VXHMCA3]], and $\\cos$ and $\\sin$ are given by their power series.\n2. At a real argument, these series are the real [[cosine|08W5RN375X]] and the real [[sine|02KTRHQMHN]]. So $\\cos 2\\pi$ and $\\sin 2\\pi$ are the values of the real functions.\n3. Only the constant terms of the series are nonzero at $0$. So $\\cos 0 = 1$ and $\\sin 0 = 0$.\n4. Since [[sine and cosine are $2\\pi$-periodic|0P4NQD746K]], $\\cos 2\\pi = \\cos 0 = 1$ and $\\sin 2\\pi = \\sin 0 = 0$.\n5. By steps 1 and 4, $e^{2 \\pi i} = 1$.\n6. Let $z \\in \\C$. By the [[product of exponentials|0WM7DJCM6G]], $e^{z + 2\\pi i} = e^z e^{2 \\pi i} = e^z \\cdot 1 = e^z$.\n\n## Idea\n\nThe exponential turns sums into products, so a shift by $2 \\pi i$ multiplies $e^z$ by $e^{2\\pi i}$. Euler's formula identifies $e^{2 \\pi i}$ with the point at angle $2\\pi$ on the unit circle, which is $1$.\n",
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