Symmetry Group Proof

Let $\alpha$ be an m-cycle permutation. Prove that $\alpha^j$ is m-cycle if and only if $gcd(m,j) = 1$ .

I have an idea for both ways but I am getting stuck on concluding either case.

If I assume the gcd is 1 then I am able to reduce down to $(\alpha^j)^s = \alpha$ where s satisfies: $sj+tm=1$ . But I struggle concluding that $\alpha^j$ is m-cycle.

If I assume $\alpha^j$ is m-cycle and that $k|m \and\ k|j$ . If I then raise $\alpha^k$ I should be able to expand the product with respect to k and conclude that k must be 1. But I fail to see a systematic way to write out $\alpha^k$ .