P-adic Norm of p-adic Number is Power of p

Theorem
Let $p$ be a prime number.

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

Let $x \in \Q_p: x \ne 0$.

Then:
 * $\exists v \in \Z: \norm x_p = p^{-v}$

Proof
By definition of the $p$-adic numbers $\Q_p$, the rational numbers $\Q$ are dense in $\Q_p$.

By the definition of a dense subset then $\map \cl \Q = \Q_p$.

By Closure of Subset of Metric Space by Convergent Sequence then:
 * there exists a sequence $\sequence {x_n} \subseteq \Q$ that converges to $x$.

That is:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = x$

By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence.

Since $\sequence {x_n}$ does not converge to $0$, by Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary then:
 * $\exists N \in \N: \forall n, m > N: \norm {x_n}_p = \norm {x_m}_p$

By definition of the $p$-adic norm on $\Q$:
 * $\exists v \in \Z: \norm {x_{N + 1} }_p = p^{-v}$

Hence:
 * $\forall n > N: \norm {x_n}_p = p^{-v}$

Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n}}$ defined by:
 * $\forall n: y_n = \norm {x_{N + n} }$

Then $\sequence {y_n}$ is the constant sequence $\tuple {p^{-v}, p^{-v}, p^{-v}, \dotsc}$.

Hence: