Coherent Sequence is Partial Sum of P-adic Expansion/Lemma

Theorem
Let $p$ be a prime number.

Let $\sequence{\alpha_n}$ be a coherent sequence.

For all $n \in \N$, let there exist a sequence $\sequence{b_{j,n}}_{0 \le j \le n}$ such that:
 * $\displaystyle \alpha_n = \sum_{j \mathop = 0}^{n} b_{j,n} p^j$
 * $\forall \, 0 \le j \le n : 0 \le b_{j,n} < p$

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

Proof
To prove this theorem, following lemma is used.

Lemma
The theorem is proved by induction:

Basis for the Induction
$n = 0$

By definition: $\alpha_0 = b_{0,0}$

So shown for basis for the induction.

Induction Hypothesis
This is our induction hypothesis:
 * $\alpha_k = \displaystyle \sum_{i \mathop = 0}^k b_{i,i} p^i$

Now we need to show true for $n=k+1$:
 * $\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i,i} p^i$

Induction Step
This is our induction step.

From Leigh.Samphier/Sandbox/Difference of Consecutive terms of Coherent Sequence:
 * $\exists c_{k + 1} \in \Z$:
 * $0 \le c_{k + 1} < p$
 * $\alpha_{k + 1} = c_{k + 1} p^{k + 1} + \alpha_k$

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 Lemma it follows that the two sequences:
 * $\sequence{b_{j,k + 1}}_{0 \le j \le k + 1}$
 * $\sequence{b_{0,0}, b_{1,1}, \dots, b_{k - 1, k - 1}, b_{k, k}, c_{k + 1}$

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