[proofplan] Write the cycle in the form $(a_1\ a_2\ \cdots\ a_k)$, where the entries $a_1,\dots,a_k$ are distinct elements of $\{1,\dots,n\}$. If $k=1$, the cycle is the identity on its support and has signature $1$. If $k\ge 2$, decompose the cycle as a product of exactly $k-1$ transpositions, then apply multiplicativity of the signature together with the fact that every transposition has signature $-1$. [/proofplan]

[step:Handle the one-cycle case] Assume first that $k=1$. A $1$-cycle fixes its unique listed element and fixes every element not listed, so $\sigma=\operatorname{id}_{\{1,\dots,n\}}$. By the definition of the signature, the identity permutation has signature $1$. Hence \begin{align*} \operatorname{sgn}(\sigma)=1=(-1)^0=(-1)^{k-1}. \end{align*} [/step]

[step:Write the cycle as a product of transpositions]Assume now that $k\ge 2$. Since $\sigma$ is a $k$-cycle, there exist distinct elements $a_1,\dots,a_k\in\{1,\dots,n\}$ such that \begin{align*} \sigma=(a_1\ a_2\ \cdots\ a_k), \end{align*} meaning that $\sigma(a_i)=a_{i+1}$ for $1\le i<k$, that $\sigma(a_k)=a_1$, and that $\sigma$ fixes every element of $\{1,\dots,n\}\setminus\{a_1,\dots,a_k\}$. For each $j\in\{2,\dots,k\}$, define $\tau_j\in S_n$ by \begin{align*} \tau_j=(a_1\ a_j). \end{align*} Thus each $\tau_j$ is a transposition. With the usual convention that products of permutations are composed from right to left, we claim that \begin{align*} \sigma=\tau_k\tau_{k-1}\cdots\tau_2. \end{align*} To verify this equality, evaluate both sides on every element of $\{1,\dots,n\}$. If $x\notin\{a_1,\dots,a_k\}$, then every $\tau_j$ fixes $x$, so the product fixes $x$, as does $\sigma$. For $x=a_1$, the rightmost transposition $\tau_2$ sends $a_1$ to $a_2$, and every remaining $\tau_j$ with $j\ge 3$ fixes $a_2$, so the product sends $a_1$ to $a_2$. For $x=a_i$ with $2\le i<k$, the transpositions $\tau_2,\dots,\tau_{i-1}$ fix $a_i$, the transposition $\tau_i$ sends $a_i$ to $a_1$, the transposition $\tau_{i+1}$ sends $a_1$ to $a_{i+1}$, and all later transpositions fix $a_{i+1}$. Hence the product sends $a_i$ to $a_{i+1}$. Finally, for $x=a_k$, the transpositions $\tau_2,\dots,\tau_{k-1}$ fix $a_k$, and $\tau_k$ sends $a_k$ to $a_1$. Therefore the product agrees with $\sigma$ on every element, proving the decomposition.[/step]

[guided]The purpose of this step is to replace one cycle by transpositions, because the signature is already known on transpositions. Since $\sigma$ is a $k$-cycle, choose distinct elements $a_1,\dots,a_k\in\{1,\dots,n\}$ such that \begin{align*} \sigma=(a_1\ a_2\ \cdots\ a_k). \end{align*} This means exactly that $\sigma(a_i)=a_{i+1}$ for $1\le i<k$, that $\sigma(a_k)=a_1$, and that every element outside $\{a_1,\dots,a_k\}$ is fixed by $\sigma$. For each $j\in\{2,\dots,k\}$, define the permutation $\tau_j\in S_n$ by \begin{align*} \tau_j=(a_1\ a_j). \end{align*} Each $\tau_j$ is a transposition because it exchanges the two distinct elements $a_1$ and $a_j$ and fixes all other elements of $\{1,\dots,n\}$. We use the usual convention that a product of permutations is composed from right to left. Under this convention, we prove \begin{align*} \sigma=\tau_k\tau_{k-1}\cdots\tau_2. \end{align*} Two permutations in $S_n$ are equal exactly when they have the same value on each element of $\{1,\dots,n\}$. If $x\notin\{a_1,\dots,a_k\}$, then every transposition $\tau_j$ fixes $x$, so the product $\tau_k\tau_{k-1}\cdots\tau_2$ fixes $x$, matching $\sigma(x)=x$. Next take $x=a_1$. The rightmost factor $\tau_2$ sends $a_1$ to $a_2$. Since $a_2$ is distinct from $a_1,a_3,\dots,a_k$, every later factor $\tau_3,\dots,\tau_k$ fixes $a_2$. Therefore \begin{align*} (\tau_k\tau_{k-1}\cdots\tau_2)(a_1)=a_2=\sigma(a_1). \end{align*} Now take $x=a_i$ with $2\le i<k$. The factors $\tau_2,\dots,\tau_{i-1}$ fix $a_i$ because none of them involves $a_i$. The factor $\tau_i$ sends $a_i$ to $a_1$. The next factor $\tau_{i+1}$ then sends $a_1$ to $a_{i+1}$. Every remaining factor $\tau_{i+2},\dots,\tau_k$ fixes $a_{i+1}$. Hence \begin{align*} (\tau_k\tau_{k-1}\cdots\tau_2)(a_i)=a_{i+1}=\sigma(a_i). \end{align*} Finally take $x=a_k$. The factors $\tau_2,\dots,\tau_{k-1}$ fix $a_k$, and the last relevant factor $\tau_k$ sends $a_k$ to $a_1$. Hence \begin{align*} (\tau_k\tau_{k-1}\cdots\tau_2)(a_k)=a_1=\sigma(a_k). \end{align*} Thus the product and $\sigma$ agree on every element of $\{1,\dots,n\}$, so \begin{align*} \sigma=\tau_k\tau_{k-1}\cdots\tau_2. \end{align*}[/guided]

[step:Apply multiplicativity of signature and evaluate each transposition] By [citetheorem:7877], the signature map is multiplicative on $S_n$. Applying this to the decomposition above gives \begin{align*} \operatorname{sgn}(\sigma)=\prod_{j=2}^k \operatorname{sgn}(\tau_j). \end{align*} Each $\tau_j$ is a transposition in $S_n$, so [citetheorem:7875] gives \begin{align*} \operatorname{sgn}(\tau_j)=-1 \end{align*} for every $j\in\{2,\dots,k\}$. Since there are exactly $k-1$ indices in $\{2,\dots,k\}$, we obtain \begin{align*} \operatorname{sgn}(\sigma)=\prod_{j=2}^k (-1)=(-1)^{k-1}. \end{align*} This is the desired formula. [/step]