Let $S_n$ be a symmetric group. Define the sign of a permutation $\sigma$ is denoted $sgn(\sigma)$ and defined as $+1$ if $\sigma$ is even and $−1$ if $\sigma$ is odd. I try to show that any the sign of $k$-cycle is $(-1)^{k-1}$.
My proof is that let $(i_1 i_2\cdots i_k)$ be a $k$-cycle.
Note that $$ (i_1 i_2\cdots i_k)=(i_1 i_2)(i_2 i_3)\cdots (i_{k-1} i_k). $$
By Theorem 5.4, (see Alternative proof that the parity of permutation is well defined?)
If a permutation $\alpha$ can be expressed as a product of an even number of $2$-cycles, then every decomposition of $\alpha$ into a product of $2$-cycles must have an even number of $2$-cycles.
this decomposition of $(i_1\cdots i_k)$ is always even or odd.
Hence $$ sgn((i_1 i_2\cdots i_k))=(-1)^{k-1}. $$
Question: Does this proof work?
You can remember that the sign of a cycle $\sigma$ of odd length $k$ is $1$ by noticing that $\sigma^k = \textrm{Id}$, of sign $1$.