 Solution

Because $k$ is a perfect field, for each $a\in k$ choose a compatible sequenceand choose arbitrary lifts $x_n\in {\mathcal{O}}_K$ of $a_n$. If $u\equiv v(\mathrm{mod}{\pi }^m)$, the binomial theorem and the fact that the residue characteristic is $p$ giveIt follows that $x_{n+1}^{p^{n+1}}\equiv x_n^{p^n}(\mathrm{mod}{\pi }^{n+1})$. Thus $(x_n^{p^n})$ is Cauchy, and completeness definesThe same congruence shows that the limit is independent of all lift choices. Taking products before passing to the limit proves $[ab]=[a][b]$, and reduction gives $[a]\equiv a(\mathrm{mod}\pi )$.

For uniqueness, let $s,t:k\to {\mathcal{O}}_K$ be two multiplicative lifts. Given $a$ and any $n$, choose $b\in k$ with $b^{p^n}=a$. Since $s(b)\equiv t(b)(\mathrm{mod}\pi )$, repeated powering yieldsCompleteness and separation force $s(a)=t(a)$. This is the unique Teichmuller lift.

For $x\in {\mathcal{O}}_K$, let $a_0$ be its residue and put $x_1=(x-[a_0])/\pi$. Repeat with $x_1,x_2,\dots$. Induction givesThe remainder tends to zero, proving the Teichmuller expansionReduction after subtracting successive partial sums also proves uniqueness of the digits.