Difference of Consecutive terms of Coherent Sequence

Theorem
Let $p$ be a prime number.

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

Then:
 * for all $n \in \N_{>0}$ there exists $c_n \in \N$ such that:
 * $0 \le c_n < p$
 * $\alpha_n - \alpha_{n - 1}  = c_n p^n$

Proof
By definition of a coherent sequence:
 * $\forall n \in \N_{>0}: \alpha_n \equiv \alpha_{n-1} \pmod {p^n}$

That is:
 * $\forall n \in \N_{>0}: \exists c_n \in \Z : \alpha_n - \alpha_{n-1} = c_n p^n$

So it remains to show that:
 * $\forall n \in \N_{>0} : 0 \le c_n < p$

Aiming for a contradiction, suppose for some $N \in \N_{>0}$:
 * $c_N \ge p$

Then:
 * $c_N p^N \ge p^{N + 1}$

By definition of a coherent sequence:
 * $\alpha_{N - 1} \ge 0$

Then:
 * $\alpha_N = c_N p^N + \alpha_{N - 1} \ge p^{N + 1}$

This contradicts the coherent sequence condition:
 * $\alpha_N \le p^{N + 1} - 1$

So:
 * $\forall n \in \N_{>0} : c_n < p$

Aiming for another contradiction, suppose for some $M \in \N_{>0}$:
 * $c_M < 0$

Then:
 * $-c_M \ge 1$

Now:
 * $c_M p^M + \alpha_{M-1} = \alpha_M \ge 0$

So:
 * $\alpha_{M-1} \ge -c_M p^M \ge p^M$

This contradicts the coherent sequence condition:
 * $\alpha_{M - 1} < p^M$

So:
 * $\forall n \in \N_{>0} : c_n \ge 0$