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  "body": "\n## Proof\n\nWe write the cycle as a product of $k - 1$ transpositions. Products in $S_n$ are composites of maps, so the right factor acts first. The sign needs $n \\geq 2$.\n\n1. Let $\\sigma = (a_1\\ a_2\\ \\dots\\ a_k)$ in [[cycle notation|0ZKPH2BD1T]], with $a_1, \\dots, a_k$ distinct. So $\\sigma(a_i) = a_{i+1}$ for $i \u003c k$, $\\sigma(a_k) = a_1$, and $\\sigma$ fixes all other points.\n2. Case $k = 1$. Then $\\sigma$ is the identity $e$. Since $e = (1\\ 2)(1\\ 2)$ is a product of two [[transpositions|02DTRYKGE8]], the [[definition of the sign|0SK90238BH]] gives $\\operatorname{sgn}(\\sigma) = (-1)^2 = 1 = (-1)^{0}$.\n3. Case $k \\geq 2$. Let $\\pi = (a_1\\ a_k)(a_1\\ a_{k-1}) \\cdots (a_1\\ a_2)$. This is a product of $k - 1$ transpositions. We show $\\pi = \\sigma$.\n4. A point $x$ outside $\\{a_1, \\dots, a_k\\}$ is fixed by each factor, so $\\pi(x) = x = \\sigma(x)$.\n5. The point $a_1$ goes first to $a_2$ under $(a_1\\ a_2)$. The later factors $(a_1\\ a_j)$ with $j \\geq 3$ fix $a_2$. So $\\pi(a_1) = a_2 = \\sigma(a_1)$.\n6. Let $2 \\leq i \\leq k - 1$. The factors $(a_1\\ a_j)$ with $j \u003c i$ fix $a_i$. Then $(a_1\\ a_i)$ sends $a_i$ to $a_1$, and $(a_1\\ a_{i+1})$ sends $a_1$ to $a_{i+1}$.\n7. The factors $(a_1\\ a_j)$ with $j \u003e i + 1$ fix $a_{i+1}$. So $\\pi(a_i) = a_{i+1} = \\sigma(a_i)$.\n8. The point $a_k$ is fixed by all factors except the last one applied, $(a_1\\ a_k)$, which sends it to $a_1$. So $\\pi(a_k) = a_1 = \\sigma(a_k)$.\n9. By steps 4 to 8, $\\pi$ and $\\sigma$ agree at every point, so $\\sigma = \\pi$ is a product of $k - 1$ transpositions.\n10. By the definition of the sign, $\\operatorname{sgn}(\\sigma) = (-1)^{k-1}$. With step 2, this holds for every $k \\geq 1$.\n\n## Idea\n\nA $k$-cycle can be built by swapping the first entry successively with the second, third, and so on up to the last. That uses $k - 1$ swaps, so the sign is $(-1)^{k-1}$. In particular, cycles of odd length are even permutations.\n",
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