P-adic Norm not Complete on Rational Numbers

Theorem
The normed vector space $(\Q,\vert\cdot\vert_p)$ does not define a complete metric space

Proof
Consider the sequence $\left(a^{p^n}\right)_{n\in\N}\subseteq \Q$ where $1\leq a\lt p$ is an integer.

Assume $n\in \N$, then:

$\left\vert a^{p^{n+1}}-a^{p^n}\right\vert_p=\left\vert a^{p^n}(a^{p^n(p-1)}-1)\right\vert_p$

From Fermat's Little Theorem we get that $a^{p^n(p-1)}-1\equiv 0\mod p^n$ so:

$\left\vert a^{p^n}(a^{p^n(p-1)}-1)\right\vert_p\leq p^{-n}\xrightarrow{n\to\infty}0$

Hence $\left(a^{p^n}\right)_{n\in\N}$ is a cauchy sequence in $(\Q,\vert\cdot\vert_p)$

On the other hand:

Denote $x_n:=a^{p^n}$ and $x=\lim_{n\to\infty}x_n$

Since $x=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}(x_n)^p=\left(\lim_{n\to\infty}x_n\right)^p=x^p$ we get that $x=x^p$.

Also from (3) on P-adic Norm is a Norm $\vert x-a\vert=\vert x-a^{p^n}+a^{p^n}-a\vert\leq\max\{\vert x-a^{p^n}\vert,\vert a^{p^n}-a\vert\}=\vert a^{p^n}-a\vert\leq \vert a^{p^n-1}-1\vert<1$

And from $\vert x-a\vert_p=\dfrac{1}{p^{\nu_p{x-a}}}<1$ follows that $p\mid (x-a)$.

Hence from $x^p=x$ follows that if $a\ne 1$ and $a\ne p-1$ then $x$ must be a non-trivial $p-1$-th root of unity in $\Q$ which is a contradiction.

In conclusion, $x\not\in\Q$