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

Suppose first that $y\ge 1$. We prove that the set $S=\{x^n\}$, $n=0,1,2,\dots$ is unbounded. Suppose to the contrary that $S$ bounded. Let $b$ be the supremum. Then for some $n$, we have $x^n \gt \frac{b}{(x+1)/2}$. Then $x^{n+1}\gt b\frac{x}{(x+1)/2}\gt b$. This contradicts the fact that $b$ is an upper bound of $S$. 
Thus the set of positive integers $k$ such that $x^k\gt y$ is non-empty. Let $n$ be the smallest element of this set. Then $x^{n-1}\le y$, and we are finished.
For $0\lt y\lt 1$, the easiest way to proceed is to show that there is a non-negative  integer $m$ such that $x^m \lt \frac{1}{y}\le x^{m+1}$. The argument for this is almost the same as the previous one. 
