If $\sigma=(a_1 a_2 ... a_n)$ and $|\sigma|$ is odd, then what is $\sigma^2$? I'm trying to understand the way to infer the power of a permutation.

If $\sigma=(a_1 a_2 ... a_n)$ is a $k$-cycle and $k=|\sigma|$ is odd, then how can I infer what $\sigma^2$ is?
HINT: Look at some examples: 

$$\begin{align*}
&amp;(1)^2=(1)\\
&amp;(123)^2=(132)\\
&amp;(12345)^2=(13524)\\
&amp;(123457)^2=(1357246)
\end{align*}$$

There’s a pretty obvious pattern developing here. We can guess, for instance, that

$$(a_1a_2a_3a_4a_5a_6a_7a_8a_9)^2=(a_1a_3a_5a_7a_9a_2a_4a_6a_8)\;.$$

Try to express this pattern in a general form and then prove that it actually holds for all cycles of odd length.
It has nothing to see with the length of the cycle:
$$\sigma^2=(a_1\,a_3\,a_5\dots a_{2i-1\bmod n}\dots a_{2n-1\bmod n}).$$