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  "body": "\n## Proof\n\n1. Let $x$ be real. Since $x$ is real, the power series of $\\cos x$ and $\\sin x$ are the real [[sine and cosine|0WB57D9488]].\nSo [[Euler's formula|04R3CY41VJ]] gives $e^{i x} = \\cos x + i \\sin x$ for every real $x$.\n2. In the series for the [[complex exponential|054VXHMCA3]] at $0$, every term with $k \\ge 1$ is $0$. So $e^0 = 1$.\n3. Claim: $(e^{i x})^n = e^{i n x}$ for each integer $n \\ge 0$ and each real $x$. We argue by induction on $n$.\n4. For $n = 0$, both sides equal $1$, by step 2.\n5. Assume the claim for $n$. By the [[product of exponentials|0WM7DJCM6G]], $(e^{i x})^{n+1} = e^{i n x} e^{i x} = e^{i (n + 1) x}$.\nThis proves the claim.\n6. By the product of exponentials and step 2, $e^{i x} e^{-i x} = e^0 = 1$. So $e^{i x} \\neq 0$ and $(e^{i x})^{-1} = e^{-i x}$.\nThus negative powers of $e^{i x}$ are defined.\n7. Let $n = -m$ with $m$ a positive integer. Then $(e^{i x})^n = \\bigl((e^{i x})^{-1}\\bigr)^m = (e^{-i x})^m$, by step 6.\n8. Step 3, applied to the real number $-x$, gives $(e^{-i x})^m = e^{-i m x} = e^{i n x}$.\nSo $(e^{i x})^n = e^{i n x}$ for every integer $n$.\n9. By step 1 for $x$ and for $n x$, $(\\cos x + i \\sin x)^n = (e^{i x})^n = e^{i n x} = \\cos(n x) + i \\sin(n x)$.\n\n## Idea\n\nEuler's formula turns $\\cos x + i \\sin x$ into $e^{i x}$.\nPowers of an exponential multiply the exponent, because exponentials turn sums into products.\nNegative powers work since $e^{-ix}$ is the inverse of $e^{ix}$.\n",
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