If $(x_n)_{n\geq 1}$, $x_n \in \mathbb{Q}$ is p-adic Cauchy, show ord$_p(x_n)$ eventually constant. Number Theory 1: Fermat's Dream asks the reader to verify the following.  They then use this to extend the definition of Ord$_p$ to $\mathbb{Q}_p$.   
Let $p$ be prime and $a\neq 0$.
If $(x_n)_{n\geq 1}$, $n \in \mathbb{N}$ is a p-adic Cauchy sequence of rationals whose class is $a \in \mathbb{Q}_p$, show that ord$_p(x_n)$ is constant for sufficiently large $n$.
 A: I explained in a comment that the other answer makes a false claim about what it means to not converge to $0$. Luckily, one can correct it, as follows:
Statement: If a sequence $(x_n)$ is Cauchy with respect to any ultrametric value $\lvert \cdot \rvert$, and $x_n$ does not converge to $0$, then there is an $n \in \mathbb N$ such that $\lvert x_n \rvert = \lvert x_{n+1} \rvert = \lvert x_{n+2} \rvert = \dots$
Proof: The sequence not converging to $0$ means there exists a certain $\epsilon_0 >0$ such that for all $k \in \mathbb N$, there is $n \ge k$ with $\lvert x_n \rvert \ge \epsilon_0$.
The sequence being Cauchy means that for all $\epsilon > 0$ there is $N \in \mathbb N$ such that for all $\ell,m \ge N$, we have $\lvert x_\ell -x_m \rvert < \epsilon$.
This applied to $\epsilon = \epsilon_0$ means there is $N \in \mathbb N$ such that for all $\ell, m \ge N$, we have $\lvert x_\ell -x_m \rvert < \epsilon_0$. Then the fixed first condition applied to $k :=N$ means there is $n\ge N$ such that, on the one hand, $$\lvert x_n \rvert \ge \epsilon_0, $$ while on the other hand, still $$\lvert x_n - x_m \rvert < \epsilon_0$$ for all $m \ge N$, in particular $m \ge n$. Combined, we have that for all $m \ge n$,
$$\lvert x_n - x_m \rvert < \epsilon_0 \le \lvert x_n \rvert \le \max (\lvert x_n \rvert, \lvert x_m \rvert) .$$
Now we use the ultra in ultrametric for the first time: The strong triangle inequality, sharpened to what is sometimes called the "ultrametric maximum principle" or similar, says that
$$\text{ If } \lvert a \rvert \neq \lvert b \rvert \text{ then } \lvert a \pm b \rvert = \max(\lvert a \rvert, \lvert b \rvert)$$
which by contraposition means
$$\text{ If } \lvert a \pm b \rvert < \max(\lvert a \rvert, \lvert b \rvert) \text{ then } \lvert a \rvert =\lvert b \rvert . $$
Apply this to $a = x_n$ and $b =x_m$ for $m = n+1, n+2, n+3, \dots$
A: If $(x_n)$ is a sequence of rational numbers that doesn't converge to zero, then the sequence of valuations, $|x_n|_p$ is constant for sufficiently large $n$.
Proof: $(x_n)$ does not tend to zero, so there exists a natural number $N$ and rational $\varepsilon > 0$ such that $n \ge N$ implies $|x_n| > \varepsilon$. $(x_n)$ fulfills the Cauchy condition for convergence, so there exists another natural number $N'$ for which the following statement is true: if $n, m \ge N'$ then $|x_n - x_m| < \varepsilon$.
Now set $M$ to be the larger of these two numbers, $N$ and $N'$ and notice, that from $n, m \ge M$ it follows that $|x_n - x_m| < \max(|x_n|, |x_m|)$. Finally the non-archimedean property gives $|x_n| = |x_m|$.
