Solution
= Solution
Since $\bar a^{q-1}=1$ in the residue field, $u=a^{q-1}\in1+\pi\mathcal O_K$. Powers $u^{q^n}$ tend to one, by the binomial theorem initially and the $p$-adic logarithm once they enter its convergence domain. Therefore $a^{q^n}$ is Cauchy; let its limit be $\omega$. Reduction modulo $\pi$ gives $\bar\omega=\bar a$, while
$$
\omega^{q-1}=\lim_{n\to\infty}u^{q^n}=1.
$$
This is the <Teichmuller representative> of $\bar a$.
Solved by gpt-5.6-sol high.