Let $K$ be complete with discrete valuation $v$, valuation ring $\mathcal{O}_K$, maximal ideal $\mathfrak{m}$, and residue field $\kappa = \mathcal{O}_K/\mathfrak{m}$.

Let $f(x) \in \mathcal{O}_K[x]$ and $a_0 \in \mathcal{O}_K$ such that $v(f(a_0)) > 0$ and $v(f'(a_0)) = 0$ (i.e., $\bar{f}(\bar{a}_0) = 0$ and $\bar{f}'(\bar{a}_0) \neq 0$ in $\kappa[x]$).

Step 1: Newton iteration

Define sequence $\{a_n\}$ by Newton's method: $a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)}$

We prove $\{a_n\}$ is Cauchy and converges to a root of $f$.

Step 2: Valuation estimates

Assume $v(a_n - a_{n-1}) \geq c_n$ with $c_n \to \infty$. Then: $f(a_{n+1}) = f(a_n) - f'(a_n) \cdot \frac{f(a_n)}{f'(a_n)} + O((f(a_n)/f'(a_n))^2)$

The main term cancels. Since $v(f'(a_n)) = 0$ remains constant and $v(f(a_n))$ increases, we get: $v(f(a_{n+1})) \geq 2v(f(a_n)) - v(f'(a_n)) = 2v(f(a_n))$

So $v(f(a_n))$ doubles at each step: if $v(f(a_0)) \geq 1$, then $v(f(a_n)) \geq 2^n$.

Step 3: Cauchy property

From $a_{n+1} - a_n = -f(a_n)/f'(a_n)$: $v(a_{n+1} - a_n) = v(f(a_n)) - v(f'(a_n)) \geq 2^n$

So $v(a_{n+1} - a_n) \to \infty$, proving $\{a_n\}$ is Cauchy.

Step 4: Convergence to root

Since $K$ is complete, $a = \lim_{n \to \infty} a_n$ exists. By continuity of $f$: $f(a) = \lim_{n \to \infty} f(a_n) = 0$

since $v(f(a_n)) \to \infty$ means $f(a_n) \to 0$.

Step 5: Uniqueness

If $b$ is another root with $\bar{b} = \bar{a}_0$, then for $c = (a+b)/2$, the mean value theorem gives: $f(a) - f(b) = f'(c)(a - b)$

Since $f(a) = f(b) = 0$ and $v(f'(c)) = v(f'(a_0)) = 0$, this implies $v(a - b) = \infty$, so $a = b$.