This is an elementary $p$-adic theory question. Granted $d(x,y)=|x-y|_q$ is a metric on $\mathbb{Q}$, and $|\cdot|_q$ is a norm such that
$$|x|_q=q^{-ord_q x}$$
where $ord_q x$ is the largest exponent $k$ such that $q^k|x$ if $x$ is an integer, and $ord_qx=ord_qa-ord_qb$ if $x\in \mathbb{Q}$, with $x=a/b$ for $a,b$ integers. Also $ord_q 0 = \infty$.
I want to show $\{p^n\}_{n \in \mathbb{N}}$ is not Cauchy in the given metric. I want to come up with a fixed $\epsilon>0$ such that for all $N\in \mathbb{N}$ we can find $m\geq n>N$ which satisfy $|p^m-p^n|_q\geq \epsilon$. The latter expression becomes $|p^n|_q|p^{m-n}-1|_q\geq\epsilon$ but since $p$ and $q$ are relatively prime, we have $|p^n|_q=1$ so I'm left with
$$|p^{m-n}-1|_q\geq\epsilon.$$ Here's where my trouble begins since I couldn't find the right $\epsilon$ and $m,n$ so that the inequality holds for all $N$. I realize that if $p^{m-n}-1 \not\equiv 0 \pmod q$ then I'm done, but I cannot find the connection. Any ideas?
Restrict to $m = n+1$. While it might happen that $p-1$ is divisible by $q$, there is a fixed largest power $q^l$ that divides $p-1$ and taking $\epsilon$ smaller than the norm of that power will do.
Possibly I should add that you cannot always find $p^{m-n} -1 $ that is not divisible by $q$ for if $q \mid (p-1 )$ then $q \mid (p-1) \mid p^k - 1 $ for all $k \ge 0$.