Let $b\in\mathbb{R}$ and $b>1$. Show that for all $r\in\mathbb{R}\exists n\in\mathbb{N}$ such that $r<b^n$

Question

So I'm working through Rudin (currently on chapter 2) and I have to prove the statement in the title and also that for any $r\in\mathbb{R}$ where $r>0$ we have that there exists $n\in\mathbb{N}$ such that $(\frac{1}{b})^n<r$. For the first statement I attempted to do a proof by contradiction by assuming a point $x$ is the supremum of the set of $b^n$ for all $n$ such that $b^n<r$ and showing that this doesn't make sense but I can't seem to make progress. Was hoping for any guidance in the right direction. Thanks in advance!

Answer 1

Way 1: Let $b=1+\delta$. By the Bernoulli Inequality (easily proved by induction) or by the Binomial Theorem, we have $$b^n=(1+\delta)^n\ge 1+n\delta.$$ Thus $n\delta\lt r$ for all $r$, contradicting the fact that $\mathbb{R}$ is Archimedean.

Way 2: We can go back to basics. If there is no such $n$, then the set $B$ of all $b^n$ is bounded above, so has a least upper bound $x$. It follows that for any $w\lt x$, there is an $n$ such that $b^n\gt w$. Let $w=\frac{x}{1+\delta/2}$.

Then $b^{n+1}\gt \frac{x}{1+\delta/2}\cdot (1+\delta)\gt x$, contradicting the fact that $x$ is an upper bound of $B$.