Theorem: P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p > 3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof:
Let $p > 3$.
Then there exists $a \in \Z: 1 < a < p-1$.
Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 < a < p-1$.
Let $n \in \N$.
Then:
:$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$
From {{Corollary|Euler's Theorem (Number Theory)|1}}:
:$a^{p^n \paren {p - 1} } - 1 \equiv 0 \pmod {p^n}$
so:
:$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$
That is:
:$\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$
By Characterisation of Cauchy Sequence in Non-Archimedean Norm
:$\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.
{{AimForCont}} $\sequence {x_n}$ converges to some $x \in \Q$.
That is:
:$x = \ds \lim_{n \mathop \to \infty} x_n$
By Modulus of Limit on a Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$
Since $\forall n, p \nmid a^{p^n} = x_n$, then:
:$ \norm {x_n }_p = 1$
So:
:$\norm x_p =  \ds \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$ 
By Axiom (N1) of a norm on a division ring then:
:$x \ne 0$.
Since:
{{begin-eqn}}
{{eqn | l = x
      | r = \lim_{n \mathop \to \infty} x_n
      | c = Definition of $x$
}}
{{eqn | r = \lim_{n \mathop \to \infty} x_{n + 1}
      | c = Limit of Subsequence equals Limit of Sequence
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n}^p
      | c = Definition of $x_n$
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} x_n}^p
      | c = Product Rule for Sequences in Normed Division Ring
}}
{{eqn | r = x^p
      | c = Definition of $x$
}}
{{end-eqn}}
and $x \ne 0$ then:
:$x^{p-1} = 1$
So:
:$x = 1$ or $x = -1$
and so $a-x$ is an integer:
:$0 < a - x < p$
It follows that:
:$p \nmid \paren {a - x}$
and so:
:$\norm {x - a}_p = 1$
Since $x_n \to x$ as $n \to \infty$ then:
:$\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$
That is:
:$\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$
Let $n > N$:
{{begin-eqn}}
{{eqn | l = \norm {x - a}_p
      | r = \norm {x - a^{p^n} + a^{p^n} - a}_p
}}
{{eqn | o = \le
      | r = \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}
      | c = P-adic Norm on Rational Numbers is Non-Archimedean Norm
}}
{{end-eqn}}
As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:
{{begin-eqn}}
{{eqn | l = \norm {x - a}_p
      | r = \norm {a^{p^n} - a}_p
      | c = Three Points in Ultrametric Space have Two Equal Distances 
}}
{{eqn | r = \norm a_p \norm {a^{p^n - 1} - 1}_p
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \norm {a^{p^n - 1} - 1}_p
      | c = as $\norm a_p = 1$
}}
{{eqn | o = <
      | r = 1
      | c = {{Corollary|Fermat's Little Theorem|4}}
}}
{{end-eqn}}
This contradicts the earlier assertion that $\norm {x - a}_p = 1$.
In conclusion:
:$\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.