Coherent Sequence is Partial Sum of P-adic Expansion

Theorem
Let $p$ be a prime number.

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

Then there exists a unique $p$-adic expansion of the form:
 * $\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$

such that:
 * $\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$

Proof
For all $n\in \N$ let:
 * $\displaystyle \alpha_n = \sum_{j \mathop = 0}^{m_n} b_{j,n} p^j$

be $\alpha_n$ written in base $p$ where:
 * $p^{m_n} \le \alpha_n < p^{m_n + 1}$
 * $\forall \, 0 \le j \le m_n : 0 \le b_{j,n} < p$

First it is shown that:
 * $\forall n \in \N : m_n \le n - 1$

By definition of a coherent sequence:
 * $\forall n \in \N: \alpha_n < p^n$

Aiming for a contradiction, suppose:
 * $\exists n \in \N: n - 1 < m_n$

Then:
 * $n \le m_n$

From Power Function on Integer between Zero and One is Strictly Decreasing:
 * $p^n \le p^{m_n} \le \alpha_n$

This contradicts the coherent sequence condition:
 * $\forall n \in \N: \alpha_n < p^n$

It follows that:
 * $\forall n \in \N : m_n \le n - 1$

Let:
 * $d_n = \begin{cases}

b_{n,n} & \text{if } m_n = n - 1\\ 0 & \text{if }m_n < n - 1 \end{cases}$

By definition:
 * $\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$

is a $p$-adic expansion.

Let:
 * $\displaystyle \beta_n = \sum_{j \mathop = 0}^{n} d_j p^j$

It remains to show that:
 * $\forall n \in \N: \alpha_n = \beta_n$

Indeed this is proved by induction:

Basis for the Induction
$n = 0$

By definition: $\alpha_0 = b_{0,0} = d_0 = \beta_0$

So shown for basis for the induction.

Induction Hypothesis
This is our induction hypothesis:
 * $\alpha_k = \beta_k$

Now we need to show true for $n=k+1$:
 * $\alpha_{k + 1} = \beta_{k + 1}$

Induction Step
This is our induction step.

By definition:
 * $\beta_{k + 1} = d_{k + 1} p^{n + 1} + \beta_k$

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

The latter sum is another basis representation of $\alpha_{k+1}$.

From Basis Representation Theorem it follows that:
 * $\forall 0 \le j \le m_k : b_{j, k} = b_{j, k + 1}$
 * $c_{k + 1} = b_{k + 1, k + 1}$

By induction hypothesis:
 * $\sum_{j \mathop = 0}^{m_k} b_{j,k} p^j = \alpha_k = \beta_k = \sum_{j \mathop = 0}^{k} d_j p^j$

From Basis Representation Theorem it follows that:
 * $\forall 0 \le j \le m_k : b_{j, k} = d_k$

Then:
 * $\forall 0 \le j \le m_k : d_k = b_{j, k + 1}$

By definition:
 * $d_{k + 1} = b_{k+1, k+1}$

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

By induction:
 * $\forall n \in \N: \alpha_n = \beta_n$

The result follows.