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  "body": "\n## Proof\n\n1. Let $n \\ge 1$, and let $\\omega_k = e^{2 \\pi i k / n}$ for $0 \\le k \u003c n$.\nBy [[the description of the nth roots of unity|057BH3JC9S]], the $n$th [[roots of unity|0S61N75H45]] are exactly the numbers $\\omega_0, \\dots, \\omega_{n-1}$.\n2. It remains to show that these $n$ numbers are distinct.\n3. Let $\\zeta = e^{2 \\pi i / n}$. By [[Euler's formula|04R3CY41VJ]] at real arguments, where the series are the real [[sine and cosine|0WB57D9488]], $\\zeta = \\cos \\frac{2 \\pi}{n} + i \\sin \\frac{2 \\pi}{n}$.\n4. By [[de Moivre's formula|03MTBVG3VE]] and Euler's formula again, $\\zeta^k = \\cos \\frac{2 \\pi k}{n} + i \\sin \\frac{2 \\pi k}{n} = \\omega_k$ for $0 \\le k \u003c n$.\n5. By [[exp(2 pi i/n) is a primitive root of unity|0CF0XGYHNJ]], $\\zeta$ is a [[primitive|0J817RTVG6]] $n$th root of unity.\nSo $\\zeta^n = 1$ and $\\zeta^j \\neq 1$ for $1 \\le j \u003c n$.\n6. Suppose that $\\omega_k = \\omega_l$ with $0 \\le l \u003c k \u003c n$. By step 4, $\\zeta^l \\zeta^{k - l} = \\zeta^l$, that is, $\\zeta^l (\\zeta^{k - l} - 1) = 0$.\n7. Since $\\zeta^n = 1$ and $0^n = 0$, we have $\\zeta \\neq 0$. So $\\zeta^l \\neq 0$, because [[$\\C$ is a field|0FRSR10DWF]] and a field has no zero divisors.\nIndeed, multiplying by the inverse of $\\zeta^l$ cancels it.\n8. So $\\zeta^{k - l} = 1$ with $1 \\le k - l \u003c n$. This contradicts step 5.\n9. Hence $\\omega_0, \\dots, \\omega_{n-1}$ are $n$ distinct numbers, and by step 1 there are exactly $n$ complex $n$th roots of unity.\n\n## Idea\n\nEvery $n$th root of unity is a power $\\zeta^k$ of $\\zeta = e^{2 \\pi i / n}$ with $0 \\le k \u003c n$.\nIf two of these powers were equal, a smaller positive power of $\\zeta$ would equal $1$, which primitivity forbids.\n",
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