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:
 * $(1) \quad m = k$
 * $(2) \quad \forall i \ge m : d_i = e_i$

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
From P-adic Number is Integer Power of p times P-adic Unit there exists $n \in \Z$:
 * $\mathbf p^n \mathbf a \in \Z^\times_p$
 * $\norm a_p = p^n$

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

Let $l = -n$.

From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
 * $d_l$ is the first index $i \ge m$ such that $d_i \neq 0$
 * $e_l$ is the first index $i \ge k$ such that $e_i \neq 0$

Now:

Similarly:

Case $l < 0$
Let $l < 0$.

Then:

Similarly:

So:
 * $m = l = k$.

Case $l \ge 0$
Let $l \ge 0$.

Then:

Similarly:

So:
 * $m = 0 = k$.

Case $m < 0$
Let $m < 0$.

From above:
 * $m = l$

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

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