P-adic Integer is Limit of Unique Coherent Sequence of Integers

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 $x \in \Z_p$.

Then there exists a unique Cauchy sequence $\sequence {\alpha_n}$:
 * $\quad\forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^n-1$
 * $\quad\forall n \in \N: \alpha_n \equiv \alpha_{n-1} \pmod {p^{n-1}}$
 * $\quad\displaystyle \lim_{n \to \infty} \alpha_n = x$

Proof
By Integers are Arbitrarily Close to P-adic Integers then for all $n \in \N$ there exists $\alpha_n \in \Z$:
 * a. $\quad0 \le \alpha_n \le p^n - 1$
 * b. $\quad\norm { x -\alpha_n}_p \le p^{-n}$

Hence the sequence $\sequence {\alpha_n}$ satisfies 1. above.

For any $n \in \N$ then:

Hence $p^{n-1} \divides \alpha_n - \alpha_{n-1}$, or equivalently, $\alpha_n \equiv \alpha_{n-1} \pmod {p^{n-1}}$

Hence the sequence $\sequence {\alpha_n}$ satisfies 2. above.

Recall that for all $n \in \N: \norm { x -\alpha_n}_p \le p^{-n}$

By Sequence of Powers of Number less than One then:
 * $\displaystyle \lim_{n \to \infty} p^{-n} = 0$

By Squeeze Theorem for Sequences of Real Numbers then:
 * $\displaystyle \lim_{n \to \infty} \norm{x - \alpha_n}_p = 0$.

Hence the limit of $\sequence{\alpha_n}$ is by definition:
 * $\displaystyle \lim_{n \to \infty} \alpha_n = x$

Hence the sequence $\sequence {\alpha_n}$ satisfies 3. above.