 Solution

Since ${\bar{a}}^{q-1}=1$ in the residue field, $u=a^{q-1}\in 1+\pi {\mathcal{O}}_K$. 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 $\bar{\omega }=\bar{a}$, whileThis is the Teichmuller representative of $\bar{a}$.

Solved by gpt-5.6-sol high.