For any $a,b \in \mathbb{Q}$, $a > 1, b> 0$, can we find an integer $n$ such that $a^n > b$? (Can we prove this without using any property of real numbers and hence functions like logarithm cannot be used because I am facing the problem in construction of real numbers)
Background
I am asking this problem because in Page 20 of Rudin's book Principles of Mathematical Analysis, the author omits details of proving some properties about multiplication between two positive real numbers and I want to complete it. In particular, I want to prove that for any positive real number $\alpha$, there exists a real number $\beta$, such that $\alpha \cdot \beta = 1^{\star}$ where $1^{\star} = \{q\in \mathbb{Q}| q < 1 \}$. I define $\beta$ as
$$\beta = \{p\in \mathbb{Q}^+ | \exists r > 1, \frac{1}{pr} \notin \alpha\} \cup \{p\in \mathbb{Q}| p \leq 0\} $$
Now, I need to prove $1^* \leq \alpha \cdot \beta$. I am thinking that for any $p \in 1^*$, $p > 0$, if we can find a $r \in \alpha$, such that $\frac{2r}{1+p} \notin \alpha$, then we can choose $s = \frac{p}{r}$ and we have $\frac{1}{s\frac{\frac{1}{p}+1}{2}} = \frac{2r}{1+p} \notin \alpha$, hence $s\in \beta$. This implies that $p = rs \in \alpha \cdot \beta$ and finish the prove. So, the problem now can be reduced to prove that there exists $r\in \alpha$ such that $\frac{2r}{1+p} \notin \alpha$ and it can be further reduced to the problem at the beginning. I feel like it is an analogy of Archimedean property on powers but I don't know how to prove it. Or there is other way to prove the existence of inverse element of multiplication? Any comment is appreciated.