How to prove $\exists n \in \mathbb{N}$ such that $a < b^n$ with $a > 0$ and $b>1$, with $a,b \in \mathbb{Q}$ A friend gave me this problem and I can't solve it. I tried to fool around with the Archimedean property but with no success. Any hints? Thanks a lot in advance!

If $b>1$, one can write $b=1+r$ for some $r>0$.
Thus, $b^m = (1+r)^m \geq 1 + mr$ for any $m\in\mathbb{N}$.
Choose $n\in\mathbb{N}$ such that $nr > a$.
We have then
$$b^n \geq 1+nr > 1 + a > a.$$
