Suppose you had the permutation $\pi= \left( \begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ 7 & 2 & 9 & 8 & 3 & 4 & 1 & 6 & 5 \end{array} \right)$.
How would you compute $\pi ^{20}$? I know that if you express a permutation as a product of disjoint cycles, such as $\pi = (1 \enspace 7)(3 \enspace 9 \enspace 5)(4 \enspace 8 \enspace 6)$, then $\pi ^{20} = (1 \enspace 7)^{20}(3 \enspace 9 \enspace 5)^{20}(4 \enspace 8 \enspace 6)^{20}$, but I do not know where top go from here. How do you multiply/square/cube/etc cycles?
Also, I've worked out that the order of $\pi$ is 6, since this is the lowest common multiple of the lengths of each of its disjoint cycles. Is this correct?
What about if you had a permutation which gave one long cycle if expressed as a product of disjoint cycles?
Since $\pi^{20} = \pi^2$: $$\pi^2 = (17)^2(395)^2(486)^2 = (359)(468)$$