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 coherent sequence $\sequence {\alpha_n}$:
 * $\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$

Proof
By the definition of a coherent sequence it needs to be proved that there exists a unique sequence $\sequence {\alpha_n}$:
 * $(1): \quad\forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^{n + 1} - 1$
 * $(2): \quad\forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
 * $(3): \quad \displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$

By Integers are Arbitrarily Close to P-adic Integers then for all $n \in \N$ there exists $\alpha_n \in \Z$:
 * $\text {(a)}: \quad 0 \le \alpha_n \le p^{n + 1} - 1$
 * $\text {(b)}: \quad \norm {x - \alpha_n}_p \le p^{-\paren{n + 1}}$

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

Lemma 1
Hence the sequence $\sequence {\alpha_n}$ satisfies $(2)$ above.

Lemma 2
Hence the sequence $\sequence {\alpha_n}$ satisfies $(3)$ above.