Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 4

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

Let $y $ be a rational $p$-adic integer.

Let $\ldots d_n \ldots d_2 d_1 d_0$ be the canonical expansion of $y$.

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

Let:
 * $\forall n \in \N: \exists A_n, r_n \in \Z$:
 * $(\text a) \quad \dfrac a b = A_n + p^{n+1} \dfrac {r_n} b$
 * $(\text b) \quad 0 \le A_n \le p^{n+1} - 1$
 * $(\text c) \quad \dfrac {a - \paren{p^{n+1} - 1} b } {p^{n+1}} \le r_n \le \dfrac a {p^{n+1}}$

Then:
 * $\forall n \in \N:$
 * $(1) \quad A_{n+1} = A_n + d_n p^{n+1}$
 * $(2) \quad r_n = d_n b + p r_{n+1}$

Proof
We have:

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

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

From Integer Coprime to all Factors is Coprime to Whole:
 * $b, p^n$ are coprime

Hence: $b \nmid p^n$

From Divisor of Product:
 * $b \divides \paren {r_n - p r_{n+1}}$

Then:
 * $\dfrac {r_n - p r_{n+1} } b \in \Z$

Hence:
 * $A_{n+1} \equiv A_n \pmod {p^{n+1}}$

By definition of coherent sequence:
 * the sequence $\sequence{A_n}$ is a coherent sequence

From Coherent Sequence is Partial Sum of P-adic Expansion:
 * the sequence $\sequence{A_n}$ is the sequence of partial sums of a $p$-adic expansion $\ds \sum_{i \mathop = 0}^n e_i p^i$