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  "body": "\n## Proof\n\nWe show that $\\zeta = e^{2 \\pi i / n}$ is a primitive $n$th root of unity.\n\n1. Let $n \\ge 1$ and $\\zeta = e^{2 \\pi i / n}$. For real arguments the power series of $\\cos$ and $\\sin$ are the real [[sine and cosine|0WB57D9488]].\nSo [[Euler's formula|04R3CY41VJ]] gives $\\zeta = \\cos \\frac{2 \\pi}{n} + i \\sin \\frac{2 \\pi}{n}$.\n2. For each integer $k \\ge 1$, [[de Moivre's formula|03MTBVG3VE]] gives $\\zeta^k = \\cos t_k + i \\sin t_k$ with $t_k = 2 \\pi k / n$.\n3. For $k = n$ we get $t_n = 2 \\pi$. Since $\\sin$ and $\\cos$ are [[$2\\pi$-periodic|0P4NQD746K]], $\\cos 2 \\pi = \\cos 0 = 1$ and $\\sin 2 \\pi = \\sin 0 = 0$.\nSo $\\zeta^n = 1$.\n4. Now let $1 \\le k \u003c n$, so $0 \u003c t_k \u003c 2 \\pi$. Assume, for a contradiction, that $\\zeta^k = 1$.\n5. By [[equality of complex numbers|05V7J4WD1C]], $\\cos t_k = 1$ and $\\sin t_k = 0$.\n6. Let $t = t_k$ if $t_k \\le \\pi$, and $t = t_k - 2 \\pi$ if $t_k \u003e \\pi$. In both cases $-\\pi \u003c t \\le \\pi$ and $t \\neq 0$.\n7. By periodicity, $\\cos t = 1$ and $\\sin t = 0$. So $1 = 1 \\cdot (\\cos t + i \\sin t)$, and $t$ is an [[argument|0ZZRSX91QR]] of $1$, since $|1| = 1$.\n8. Also $0$ is an argument of $1$ in $(-\\pi, \\pi]$, since $\\cos 0 = 1$ and $\\sin 0 = 0$.\n9. The [[principal value of the argument|0SQSQP748R]] is unique, so $t = 0$. This contradicts step 6.\n10. So $\\zeta^k \\neq 1$ for $1 \\le k \u003c n$, and $\\zeta^n = 1$ by step 3.\nBy the definition of a [[primitive root of unity|0J817RTVG6]], $\\zeta$ is a primitive $n$th root of unity.\n\n## Idea\n\nThe powers $\\zeta^k$ are the points of the unit circle at angles $2 \\pi k / n$.\nFor $1 \\le k \u003c n$ the angle lies strictly between $0$ and $2 \\pi$, so the point is not $1$.\nThe first power that returns to $1$ is the $n$th.\n",
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