 Solution

The structural facts used about a non-Archimedean local field are that it is complete for a normalized discrete valuation $v:K^{\times }\to \mathbb{Z}$, its valuation ring ${\mathcal{O}}_K=\{x:v(x)\ge 0\}$ is a discrete valuation ring, its maximal ideal is ${\mathfrak{m}}_K=\pi {\mathcal{O}}_K$ for a uniformizer $\pi$, and its residue field is a finite field $k={\mathbb{F}}_q$. Polynomial operations are continuous for this valuation topology.

A simple-root version of Hensel lemma is: if $f\in {\mathcal{O}}_K[X]$, $a_1\in {\mathcal{O}}_K$, $f(a_1)\in {\mathfrak{m}}_K$, and $f^{\prime }(a_1)\notin {\mathfrak{m}}_K$, then there is a unique $a\in {\mathcal{O}}_K$ with $f(a)=0$ and $a\equiv a_1(\mathrm{mod}{\mathfrak{m}}_K)$.

Construct $a_n$ with $f(a_n)\in {\pi }^n{\mathcal{O}}_K$. Given $a_n$, put $a_{n+1}=a_n+{\pi }^n{b}_n$. Taylor expansion over the valuation ring givesThe omitted terms contain ${\pi }^{2n}$, and $2n\ge n+1$ for $n\ge 1$. The residue of $f^{\prime }(a_n)$ remains the nonzero residue of $f^{\prime }(a_1)$, so choose $b_n$ withThis makes $f(a_{n+1})\in {\pi }^{n+1}{\mathcal{O}}_K$. The congruences $a_{n+1}\equiv a_n(\mathrm{mod}{\pi }^n)$ make $(a_n)$ a Cauchy sequence; completeness gives its limit $a\in {\mathcal{O}}_K$. Continuity and $v(f(a_n))\ge n$ give $f(a)=0$, and the first congruence gives its specified residue class.

For uniqueness, if $a,b$ are two roots in that class, polynomial division of their difference givesThe second factor is a unit, since $f^{\prime }(b)$ is a unit and $a-b\in {\mathfrak{m}}_K$. Thus $a=b$. This proves both existence and uniqueness in the stated Hensel lemma.