Let $\sigma$ be an $r$-cycle and let $k \in\Bbb N$. Let $d=\gcd(r,k)$. Write $r'=r/d$ and $k'=k/d$. Prove that $\sigma^k$ is a product of $d$ disjoint cycles each of length $r'$.
I think $ord(\sigma^k)=lcm(k,r)=r'k'd$, but I don't know if that helps. I know intuitively that the statement I need to prove is true, but I have no idea how to show it!
We can assume that $\sigma$ acts on $\{1, \dotsc, r\}$.
Since $\sigma$ is a $r$-cycle, none of the permutations $\sigma^i, i=1, \dotsc, r-1$ stabilizes some element of $\{1, \dotsc, r\}$.
So if we consisder the action of $\langle \sigma^k \rangle$ on $\{1, \dotsc, r\}$, the stabilizer of each element is trivial, hence by the orbit-stabilizer theorem the length of the orbit of every element is equal to the order of $\langle \sigma^k \rangle$, which is precisely $r'$. That's the assertion