Given that $$ a^\infty = \{ a_i\} _{i = 0}^{\infty} = (a_0a_1a_2\ldots) $$ is a peroidic sequence over $\mathrm{GF}(q)$ and its period is $p$. Let $s$ a positive interger and define $$ a^\infty(s) = \{ a_{is}\} _{i = 0}^{\infty} = (a_0a_sa_{2s}\ldots) $$ The question is to prove that if $\mathrm{gcd}(p, s) = 1$ then the period of $a^\infty(s)$ is $p$ as well.
I have worked out that the period of $a^\infty(s)$ must be a factor of $p$ since $(a_0, a_s, a_{2s}, \ldots, a_{(p-1)s})$ are the permutation of $(a_0, a_1, a_{2}, \ldots, a_{(p-1)})$. That is, $a_{(p+i)s} = a_{is}$. However, I have no idea how to prove that the period of $a^\infty(s)$ is $p$ exactly.