 Solution

Write the finite residue field as $k={\mathbb{F}}_q$, with $q=p^f$ and residue characteristic $p$. Every $\bar{a}\in k$ satisfies ${\bar{a}}^q=\bar{a}$. The polynomial $F(X)=X^q-X$ has derivative $q{X}^{q-1}-1$, whose reduction is $-1$. Hence Hensel lemma gives a unique lift $[\bar{a}]\in {\mathcal{O}}_K$ such thatThis includes $[0]=0$ and $[1]=1$. If $\bar{a}\ne 0$, its lift is a unit and satisfies $[\bar{a}{]}^{q-1}=1$.

The product $[\bar{a}][\bar{b}]$ also satisfies $X^q=X$ and has residue $\bar{a}\bar{b}$. Uniqueness therefore gives $[\bar{a}\bar{b}]=[\bar{a}][\bar{b}]$. Reduction is a left inverse of this map, so the map is injective. Restricting to nonzero residues proves the requested multiplicative Teichmuller lift:Its image is the group of $(q-1)$st roots of unity in $K$. Indeed a root of unity of that order is a unit satisfying $X^q=X$, hence is the unique lift of its residue. The only finite-field fact used here is $a^q=a$; for $a\ne 0$ it follows from Lagrange's theorem in the multiplicative group of order $q-1$.