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

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

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

From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion, $\mathbf a$ 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 $\norm{\mathbf a}_p > 1$.

Let $\textbf p$ denote the equivalence class that is identified with the prime number $p$.

By definition of the $p$-adic numbers as a quotient of Cauchy sequences:
 * the constant sequence $\tuple {p, p, p, \dots}$ represents $\mathbf p \in \Q_p$.

From P-adic Number is Integer Power of p times P-adic Unit,
 * $\exists m \in \Z: \mathbf p^m \textbf a \in \Z_p^\times$

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

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

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

From P-adic Unit has Norm Equal to One:
 * $\norm{\mathbf p^m \textbf  a}_p = 1 = p^0$

From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
 * $d_0 \neq 0$

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

By definition of the $p$-adic numbers as a quotient of Cauchy sequences:
 * the constant sequence $\tuple {p^m, p^m, p^m, \dots}$ represents $\mathbf p^m \in \Q_p$.

By definition of the product in a quotient ring:
 * the Cauchy sequence $\sequence{p^m x_n}$ is a representative of $\mathbf p^m \mathbf  a \in \Q_p$.

From Representatives of same P-adic Number iff Difference is Null Sequence:
 * $\sequence{p^m x_n - \displaystyle \sum_{i \mathop = 0}^n d_i p^i}$ is a null sequence

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

From Representatives of same P-adic Number iff Difference is Null Sequence:
 * $\sequence{\displaystyle p^{-m} \sum_{i \mathop = 0}^\infty d_i p^i}$ is a representative of $\mathbf a$.

That is:
 * $\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a representative of $\mathbf a$

where:
 * $\forall i \ge -m: e_i = d_{i + m}$

By definition of a $p$-adic expansion;
 * $\forall i \ge 0: 0 \le d_i < p$

Then:
 * $\forall i \ge -m: 0 \le e_i = d_{i + m} < p$

Since $d_0 \neq 0$, then $e_{-m} = d_0 \neq 0$.

By definition of a $p$-adic expansion;
 * $\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a $p$-adic expansion that represents $\mathbf a$

From Leigh.Samphier/Sandbox/P-adic Expansion Representative of P-adic Number is Unique:
 * $\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is the only $p$-adic expansion that represents $\mathbf a$