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  "body": "\n## Proof\n\nWe show that multiplying the sum by $\\zeta$ does not change it.\n\n1. Let $n \\ge 2$ be an integer and let $\\zeta$ be a [[primitive $n$th root of unity|0J817RTVG6]].\nBy definition, $\\zeta^n = 1$ and $\\zeta^k \\neq 1$ for each integer $k$ with $1 \\le k \u003c n$.\n2. Since $n \\ge 2$, the integer $k = 1$ satisfies $1 \\le k \u003c n$. So step 1 gives $\\zeta \\neq 1$.\n3. Let $S = \\sum_{k=0}^{n-1} \\zeta^k = 1 + \\zeta + \\dots + \\zeta^{n-1}$.\n4. Multiplication in $\\C$ distributes over addition, since [[$\\C$ is a field|0FRSR10DWF]]. So $\\zeta S = \\sum_{k=0}^{n-1} \\zeta^{k+1} = \\zeta + \\zeta^2 + \\dots + \\zeta^{n}$.\n5. By step 1, the last term is $\\zeta^n = 1$. So $\\zeta S = 1 + \\zeta + \\dots + \\zeta^{n-1} = S$.\n6. Subtracting $S$ from both sides of step 5 gives $(\\zeta - 1) S = 0$.\n7. By step 2, $\\zeta - 1 \\neq 0$. Since $\\C$ is a field, $\\zeta - 1$ has an inverse $(\\zeta - 1)^{-1}$.\n8. Multiplying step 6 by $(\\zeta - 1)^{-1}$ gives $S = 0$.\nSo $1 + \\zeta + \\zeta^2 + \\dots + \\zeta^{n-1} = 0$.\n\n## Idea\n\nMultiplying the sum by $\\zeta$ shifts every power up by one.\nThe top power $\\zeta^n$ equals $1$ and takes the place of the bottom term, so the sum is unchanged.\nA number fixed by multiplication by some $\\zeta \\neq 1$ must be $0$.\n",
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