P-adic Number has Unique P-adic Expansion Representative

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.

Let $a$ be an equivalence class in $\Q_p$.

Then $a$ has exactly one representative that is a $p$-adic expansion.

Case 1
Let $\norm{a}_p \le 1$.

From Leigh.Samphier/Sandbox/Equivalence Class in P-adic Integers Contains Unique P-adic Expansion $a$ has exactly one representative that is a $p$-adic expansion of the form:
 * $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

Case 2
Let $\norm{a}_p > 1$.

From Leigh.Samphier/Sandbox/P-adic Number is Integer Power of p times P-adic Unit,
 * $\exists m \in \Z: p^m a \in \Z_p^\times$

where $\Z_p^\times$ denotes the $p$-adic units.

By definition a $p$-adic unit is a $p$-adic integer.

From Leigh.Samphier/Sandbox/Equivalence Class in P-adic Integers Contains Unique P-adic Expansion $p^m a$ has exactly one representative that is a $p$-adic expansion of the form:
 * $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$

Let $\sequence{x_n}$ be a Cauchy sequence of rational numbers that represents $a$.

$\dots$

Then:
 * $\sequence{p^m x_n}$ is a representative of $p^m a$.

From :
 * $\sequence{p^m x_n - \displaystyle \sum_{i \mathop = 0}^n d_i p^i}$ is a Definition:Null Sequence

From Multiple Rule for Cauchy Sequences in Normed Division Rings:
 * $\sequence{x_n - \displaystyle p^{-m} \sum_{i \mathop = 0}^n d_i p^i}$ is a Definition:Null Sequence

From :
 * $\sequence{\displaystyle p^{-m} \sum_{i \mathop = 0}^n d_i p^i}$ is a representative of $a$.

That is:
 * $\sequence{\displaystyle \sum_{i \mathop = 0}^n d_i p^{i-m}}$ is a representative of $a$.