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.

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

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

Let $l$ be the index of the first non-zero coefficient in the $p$-adic expansion:
 * $l = \min \set {i: i \ge m \land d_i \ne 0}$

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

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

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

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

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

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

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

Finally:

The result follows.