P-adic Expansion Representative of P-adic Number is Unique

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 $\Q_p$.

Let $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $\mathbf a$.

Then:
 * $m = k$
 * $\forall i \ge m : d_i = e_i$

That is, the $p$-adic expansions $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ are identical.

Proof m = k
Let $\norm{\mathbf a}_p = p^{-l}$.

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

If $l \ge 0$ then by defintion of P-adic Expansion:
 * $m = 0$
 * $k = 0$

So $m = k$.

If $l < 0$ then by definition of P-adic Expansion:
 * $m < 0 \implies d_m \neq 0 \implies m = l$
 * $k < 0 \implies d_k \neq 0 \implies k = l$

So $m = k$.

Proof $d_i = e_i$ for all $i \ge m$
From P-adic Number is Integer Power of p times P-adic Unit:
 * $\exists k \in \Z : \mathbf p^k \mathbf a \in \Z^\times_p$

where $\Z^\times_p$ is the set of $p$-adic units.

From Multiple Rule for Cauchy Sequences in Normed Division Ring:
 * $\displaystyle p^k \sum_{i \mathop = m}^\infty d_i p^i = \sum_{i \mathop = m}^\infty d_i p^{i+k} = \sum_{i \mathop = 0}^\infty d_{i-m} p^i$
 * $\displaystyle p^k \sum_{i \mathop = m}^\infty e_i p^i = \sum_{i \mathop = m}^\infty e_i p^{i+k} = \sum_{i \mathop = 0}^\infty e_{i-m} p^i$

are representatives of $\mathbf p^k \mathbf a$.