[proofplan] Write $\sigma$ as the product of its pairwise disjoint nontrivial cycles. Each cycle contributes the sign dictated by its length, namely $(-1)^{k_i-1}$. Multiplicativity of the signature then turns the signature of the product into the product of the signatures. Finally, the product of powers of $-1$ is collected into a single exponent, with the identity case handled by the empty product convention. [/proofplan] [step:Write the permutation as the product of its nontrivial disjoint cycles] By hypothesis, the [disjoint cycle decomposition](/theorems/775) of $\sigma$ has nontrivial cycles $c_1,\dots,c_r$, and $c_i$ has length $k_i$ for each $i\in\{1,\dots,r\}$. Thus \begin{align*} \sigma=c_1c_2\cdots c_r. \end{align*} If $r=0$, this product is the empty product in $S_n$, so $\sigma=e_{S_n}$, where $e_{S_n}:\{1,\dots,n\}\to\{1,\dots,n\}$ denotes the identity map $a\mapsto a$. [/step] [step:Compute the signature of each cycle from its length] For each $i\in\{1,\dots,r\}$, the permutation $c_i\in S_n$ is a $k_i$-cycle. By [citetheorem:7878], \begin{align*} \operatorname{sgn}(c_i)=(-1)^{k_i-1}. \end{align*} [guided] Fix an index $i\in\{1,\dots,r\}$. The symbol $c_i$ denotes one of the nontrivial cycles appearing in the disjoint cycle decomposition of $\sigma$, and $k_i$ is its length. Therefore $c_i$ satisfies the hypotheses of [citetheorem:7878]: it is a $k_i$-cycle in $S_n$, with $1\le k_i\le n$. Applying that theorem gives \begin{align*} \operatorname{sgn}(c_i)=(-1)^{k_i-1}. \end{align*} This is the local contribution of the cycle $c_i$ to the sign of the whole permutation. The rest of the proof is to multiply these local contributions together. [/guided] [/step] [step:Use multiplicativity of the signature on the cycle product] If $r\ge 1$, repeated application of [citetheorem:7877] to the product $c_1c_2\cdots c_r$ gives \begin{align*} \operatorname{sgn}(\sigma)=\prod_{i=1}^r \operatorname{sgn}(c_i). \end{align*} Substituting the cycle computation from the previous step yields \begin{align*} \operatorname{sgn}(\sigma)=\prod_{i=1}^r(-1)^{k_i-1}. \end{align*} If $r=0$, then $\sigma=e_{S_n}$. Since $\operatorname{sgn}:S_n\to\{1,-1\}$ is a [group homomorphism](/page/Group%20Homomorphism) by [citetheorem:7877], it sends the identity of $S_n$ to the identity of $\{1,-1\}$, so $\operatorname{sgn}(\sigma)=1$, which equals the empty product. [/step] [step:Combine the powers of $-1$] Using the exponent law for integer powers in the multiplicative group $\{1,-1\}$, \begin{align*} \prod_{i=1}^r(-1)^{k_i-1}=(-1)^{\sum_{i=1}^r(k_i-1)}. \end{align*} For $r=0$, this reads $1=(-1)^0$, using the empty product and empty sum conventions. Combining this identity with the previous step proves \begin{align*} \operatorname{sgn}(\sigma)=\prod_{i=1}^r(-1)^{k_i-1}=(-1)^{\sum_{i=1}^r(k_i-1)}. \end{align*} [/step]