Parity of K-Cycle

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

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

Thus $$\sgn \left({\pi}\right) = \left({-1}\right)^{k-1}$$ or $$\sgn \left({\pi}\right) = \left({-1}\right)^{k+1}$$.

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

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

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