P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient

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 $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ be a $p$-adic expansion.

Let $\mathbf a$ be the equivalence class in $\Q_p$ containing $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$.

Let $l$ be the index of the first nonzero coefficient in the $p$-adic expansion.

That is, $l$ is the first index in the ordered set $\set {i: i \ge m \land d_i \ne 0}$.

Then:
 * $\norm {\mathbf a}_p = p^{-l}$

Proof
For all $n \ge m$, let:
 * $\alpha_n = \displaystyle \sum_{i \mathop = m}^n d_i p^i$

By assumption:
 * $\sequence{\alpha_n}$ is a representative of $\mathbf a$

By definition of the induced norm:
 * $\norm{\mathbf a}_p = \displaystyle \lim_{n \mathop \to \infty} \norm{\alpha_n}_p$

From Eventually Constant Sequence Converges to Constant it is sufficient to show:
 * $\forall n \ge l + 1 : \norm{\alpha_n}_p = p^{-l}$

Let $n \ge l + 1$.

Then:

The sum $\displaystyle \sum_{i \mathop = l+1}^n d_i p^i$ can be rewritten:
 * $\displaystyle \sum_{i \mathop = l+1}^n d_i p^i = p^{l + 1} \sum_{i \mathop = l+1}^n d_i p^{i - \paren{l + 1} }$

By definition of divisor:
 * $\displaystyle p^{l+1} \divides \sum_{i \mathop = l+1}^n d_i p^i$

Then:

By definition of a $p$-adic expansion:
 * $0 < d_l < p$

Then:
 * $p^l \divides d_l p^l$
 * $p^{l + 1} \nmid d_l p^l$

By definition of $p$-adic norm:
 * $\norm{ d_l p^l }_p = p^{-l}$

Thus:
 * $\displaystyle \norm{\sum_{i \mathop = l+1}^n d_i p^i}_p < p^{-l} = \norm{ d_l p^l }_p$

Finally:

The result follows.