Is it true that for every $p$ and $q$ prime there exist infinitely many $k$ such that $p^k - 1$ is divisible by $q$?

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This question is interesting because it helps to ask for another question:

What is multiplicative group of algebraic closure $\bar{\mathbb{F}}_p^*$ of $\mathbb{F}_p$?

I know that for algebraic extension $\mathbb{F}_q \supset \mathbb{F}_p$, it's true that $(\mathbb{F}_q)^* =$ ciclic group on $q - 1$ element.

If the answer for question in title is yes, then we know a lot of subgroups of $\bar{\mathbb{F}}_p^*$. More presicely we know, that for every n there exists subgroup $\mathbb{Z} /n\mathbb{Z} \subset \bar{\mathbb{F}}_p^*$ .

I know that $\widehat{\mathbb{Z}}$ has the same property. If I'm right that $\widehat{\mathbb{Z}} = \times_r \mathbb{Z}_r$, where $r$ is prime, and $\mathbb{Z}_r$ is $r$-adic group.

Is it true that $\bar{\mathbb{F}} _p^* =\widehat{\mathbb{Z}}$?