Canonical P-adic Expansion of Rational is Eventually Periodic/Necessary Condition

Theorem
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.

Let $x$ be a rational number.

Then:
 * the canonical expansion of $x$ is eventually periodic

Proof
Let $x$ be a rational number.

Let $\ldots d_n \ldots d_2 d_1 d_0. d_{-1} d_{-2} \ldots d_{-m}$ be the canonical expansion of $x$.

It is sufficient to show that the canonical expansion $\ldots d_n \ldots d_2 d_1 d_0$ is eventually periodic.

Let $y$ be the $p$-adic number with canonical expansion:
 * $\ldots d_n \ldots d_2 d_1 d_0$

We have:
 * $y = x - \ds \sum_{i \mathop = -m}^{-1} d_i p^i$

So:
 * $y$ is a rational number

By definition of $p$-adic integer:
 * $y$ is a $p$-adic integer

Let:
 * $y = \dfrac a b : a \in \Z, b \in Z_{> 0}$ are coprime

From Characterization of Rational P-adic Integer:
 * $p \nmid b$

From Prime not Divisor implies Coprime:
 * $b, p$ are coprime

Lemma 3
From lemma 3:
 * the sequence $\sequence{r_n}$ takes only finitely many values

Hence:
 * $\exists m, l \in \N : l > 0 : r_m = r_{m+l}$

From lemma 2:

As $b, p$ are coprime:
 * $p \nmid b$

From Divisor of Product:
 * $p \divides \paren {d_m - d_{m + l} }$

By definition of canonical expansion:
 * $d_m, d_{m + l} \in \set{0, 1, \ldots, p-1}$

Hence:
 * $d_m = d_{m+l}$

We have:

Repeating this argument:
 * $\forall n \ge m: r_{n + 1} = r_{n + l + 1}$ and $d_{n + 1} = d_{n + l + 1}$

It follows that $\ldots d_n \ldots d_2 d_1 d_0$ is eventually periodic.