Parity of K-Cycle

Theorem
Let $\pi$ be a $k$-cycle.

Then $\operatorname{sgn} \left({\pi}\right) = \begin{cases} 1 & : k \ \text {odd} \\ -1 & : k \ \text {even} \end{cases}$.

Thus:
 * $\operatorname{sgn} \left({\pi}\right) = \left({-1}\right)^{k - 1}$

or equivalently:
 * $\operatorname{sgn} \left({\pi}\right) = \left({-1}\right)^{k + 1}$

Proof
From Transposition is of Odd Parity, any transposition is of odd parity.

From K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $k - 1$ transpositions.

Thus $\pi$ is even $k$ is odd.