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 $\set{i : i \ge m \land d_i \neq 0}$.

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

Proof
For all $n \in \N$:
 * 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:
 * $\forall n \ge l : \norm{\alpha_n}_p = p^{-l}$

The is proved by induction:

Basis for the Induction
$n = l$

Then:

Induction Hypothesis
This is our induction hypothesis:
 * $\alpha_k = p^{-l}$

Now we need to show true for $n=k+1$:
 * $\alpha_{k + 1} = p^{-l}$

Induction Step
This is our induction step.

From Difference of Consecutive terms of Coherent Sequence there exists $c_{k + 1} \in \Z$:
 * $(a) \quad 0 \le c_{k + 1} < p$
 * $(b) \quad \alpha_{k + 1} = c_{k + 1} p^{k + 1} + \alpha_k$

From the induction hypothesis:
 * $\alpha_{k + 1} = c_{k + 1} p^{k + 1} + \sum_{j \mathop = 0}^k b_{j,j} p^j$

From the Zero Padded Basis Representation it follows that the two sequences:
 * $\sequence{b_{j,k + 1}}_{0 \le j \le k + 1}$

and
 * $\sequence{b_{0,0}, b_{1,1}, \dots, b_{k - 1, k - 1}, b_{k, k}, c_{k + 1}}$

are equal.

Then:
 * $\forall j \in \closedint 0 {k + 1} : b_{j, k + 1} = b_{j, j}$

It follows that:
 * $\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i,i} p^i$

By induction:
 * $\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n b_{i,i} p^i$