Any power of a prime-length cycle is a cycle

The claim is slightly incorrect. The correct claim is:

The permutation $\beta = \alpha^k$ is a cycle if and only if $s$ does not divide $k$.

Clearly, if $s$ divides $k$, then $\alpha^k$ is just the identity permutation, and hence not a cycle. Now for the other direction, assume $s$ does not divide $k$. It suffices to show that for each $t \in \{ 1, 2, \ldots, s \}$, there exists an $i$ such that $\beta^{i}(a_1) = a_t$; i.e., $\alpha^{ki}(a_1) = a_t$. But this happens if and only if $$ ki \equiv t-1 \pmod s. $$

Assuming $s$ does not divide $k$, since $s$ is a prime, $k$ has a multiplicative inverse $k^{-1}$ modulo $s$. Now picking $i = k^{-1} \cdot (t-1)$ gives what we want.