Parity_of_K_Cycle_v2

Let $\pi$ be a $k$-cycle. Then $\map \sgn \pi = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$ Thus $\map \sgn \pi = \paren {-1}^{k - 1}$ or equivalently $\map \sgn \pi = \paren {-1}^{k + 1}$