Multiplicative version of the principle of Archimedes 
Any clear proof of the above theorem is greatly appreciated.

The set $S= \{x^i\ \vert\ i \in \mathbb{Z}\}$ has a greatest lower bound $c$. It can be easily shown that $0\leq c$.
If $c$ is a lower bound of $S$, so is $xc$, and hence $xc \leq c$ and $(x-1)c\leq 0$. Since $x-1>0$ and $c$ is nonnegative, it follows that $c=0$.
Now $y>0=c$ so it is not a lower bound of $S$, and $x^{n} < y$ for some $n$ (*). 
Similarly $x^{n} < y^{-1}$ for some integer $n$, hence $y < x^{-n}$ for some $n$. (**)
Finally let $T$ be the set of integers $m$ for which $x^m > y$. $T$ is nonempty (**), bounded below (*), and hence has a least element $k$. This $k$ satisfies
$$x^{k-1} \leq y < x^{k},$$ 
as desired.
