In a proof of Hilbert 90 we want to show that $Z^1(G,\overline{\mathbb F_q}) = B^1(G,\overline{\mathbb F_q})$, where $G = <Fr_q>$ is the cyclic group of Frobenius endomorphisms (with composition), and $\overline{\mathbb F_q}$ is the algebraic closure of the finite field $\mathbb F_q$ with $q=p^m$
The "$\supseteq$" is trivial, so we want to show that each 1-cocycle $\lambda:G \to \overline{\mathbb F_q}$ is also a 1-coboundary.
For an easier notation let $f := Fr_q$.
For some $k \in \mathbb Z, k \geq 1 $ we get $\lambda(f^k) = \lambda(f) + f\lambda(f)+\ldots + f^{k-1}\lambda(f) = \lambda(f) + \lambda(f)^q + \ldots + \lambda(f)^{q^{k-1}}$. (Where $f^k = \underbrace{f \circ f \circ \ldots \circ f}_{k-\text{times}}$)
The following statement is what I do not understand:
Because there are only finitely many powers $\lambda(f)^{q^i}$ and because $\overline{\mathbb F_q}$ has positive characteristic, there are only finitely many possibilities for $\lambda(f^k) \in \overline{\mathbb F_q}$.
Why are there only finitely man powers $\lambda(f)^{q^i}$? And how does that imply that there are only finitely many possibilities for $\lambda(f^k)$?
Note that $\displaystyle \overline{\Bbb F_q} = \bigcup_{n\in\Bbb N} \Bbb F_{q^n}$ so that $\exists n\in\Bbb N$ such that $\lambda(f)\in \Bbb F_{q^n}$ : hence $\lambda(f)^{q^n}=\lambda(f)$ and $\lambda(f)$ has finitely many distinct powers.
Now, since the field is of characteristic $p$, for a $k$ large enough, $\lambda(f^k)$ is a sum containing $p$ times a same power of $\lambda(f)$ (because we have finitely many powers, as said above), and we can simplify. In fact, a sum of powers of $\lambda(f)$ is always obtained by taking finitely many times (up to simplifying by $p$ times something) some powers of $\lambda(f)$ (the number of which being finite). Thus there is only finitely many possibilities for such a sum.