Let $k'/k$ be (maximal) unramified local field extension with Frobenius automorphism $f:x\to x^q$ where $q$ is cardinality of residue field of $k$. Let $a_1,a_2$ be two solutions of equation $x-\gamma x^q=\beta$ where $b\in O_{k'}$ and $v(\gamma)\geq 1$, $v$ is valuation extended from $k$ to $k'$ and $O_{k'}$ is valuation ring associated to extended valuation $v$.
"..., then $a_1-a_2=\gamma(a_1^q-a_2^q), v(a_1-a_2)=v(f(a_1-a_2))=v(f(a_1)-f(a_2))$,..."
$\textbf{Q1:}$ I do not see how $v(f(a_1-a_2))=v(f(a_1)-f(a_2))$ follows from $a_1,a_2$ being solutions of equation $x-\gamma f(x)=\beta$. The whole point of above argument is to conclude that $a_1=a_2$ for uniqueness of solution. The rough expectation from me is that $q$ is in prime ideal. Denote $f(a_1-a_2)=f(a_1)-f(a_2)+R$ where $R$ is remainder terms. $v(R)\geq 1$ by $q$ inside prime ideal. However, I need to compare $v(f(a_1)-f(a_2)),v(R)$ to conclude the equality $v(f(a_1-a_2))=v(f(a_1)-f(a_2))$.
I did not use above machinery to conclude $a_1=a_2$. Instead, I used $v(a_1-a_2)=v((a_1-a_2)^q)$ by unramified extension.(Note that $x\to x^q$ belongs to element fixing ground field $k$.) So $v((a_1-a_2)^q)=v(\gamma^q(a_1^q-a_2^q)^q)$. Now factorization of $v((a_1^q-a_2^q))\geq v(a_1-a_2)$ to bring another $v(\gamma^q)$ out. This procedure can be iterated indefinitely. Thus $v(a_1-a_2)=\infty$.
$\textbf{Q2:}$ Do I need "maximal" unramified extension here?(i.e. Do I need maximality?)
Ref. K. Iwasawa Local Class Field Theory Chpt 3, Prop 3.12.