P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $\Z_p$ be the $p$-adic integers.

Let $x \in \Z_p$.

Let $\sequence {\alpha_n}$ be a sequence such that:
 * $\quad\forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^n-1$
 * $\quad\forall n \in \N: \alpha_n \equiv \alpha_{n-1} \pmod {p^{n-1}}$
 * $\quad\displaystyle \lim_{n \to \infty} \alpha_n = x$

Then $\sequence {\alpha_n}$ is a unique sequence satisfying properties 1., 2. and 3. above.

Proof
Suppose that there exists a sequence $\sequence {\alpha'_n}$ with:
 * 1'. $\quad\forall n \in \N: \alpha'_n \in \Z$ and $0 \le \alpha'_n \le p^n-1$
 * 2'. $\quad\forall n \in \N: \alpha'_n \equiv \alpha'_{n-1} \pmod {p^{n-1}}$
 * 3'. $\quad\displaystyle \lim_{n \to \infty} \alpha'_n = x$

Aiming for a contradiction, suppose:
 * $\alpha'_N \neq \alpha_N$ for some $N \in \N$

By Initial Segment of Natural Numbers forms Complete Residue System then:
 * $\alpha'_N \not \equiv \alpha_N \pmod {p^N}$.

Then for all $n \gt N$:

That is, for all $n \gt N$:
 * $\alpha'_n \not \equiv \alpha_n \pmod {p^N}$

Hence for all $n \gt N$:
 * $\norm{\alpha'_n - \alpha_n}_p \gt p^{-N}$

By 3. the limit of $\sequence{\alpha_n}$ is $x$ then:
 * $\exists N_1 \in \N: \forall n \ge N_1: \norm{x - \alpha_n}_p \le p^{-N}$

Similarly for $\sequence{\alpha'_n}$:
 * $\exists N_2 \in \N: \forall n \ge N_2: \norm{x - \alpha'_n}_p \le p^{-N}$

Let $M = \max \set {N+1, N_1, N_2}$.

Then:
 * $\norm{\alpha'_M - \alpha_M}_p \gt p^{-N}$
 * $\norm{x - \alpha_M} _p\le p^{-N}$
 * $\norm{x - \alpha'_M}_p \le p^{-N}$

But:

This contradicts the previous assertion that $\norm{\alpha'_M - \alpha_M}_p \gt p^{-N}$

Hence $\sequence{\alpha'_n} = \sequence{\alpha_n}$.

The result follows.