I'm reading Courant's Integral and Differential Calculus, here:
He says that $a^{r_m}$ remains bounded, but bounded by what? I thought about two possibilities:
Bounded by $a^{a}$, but at this point, we still don't know this limit exists.
Bounded by $a^{r_n}$ because $r_n > r_m$ and as $a>1$, then we must have $a^{r_n} > a^{r_m}$. But I guess that this have the same problem of the last item because if we still don't know that the limit exists, then it could be possible that $a^{r_n} \to \infty$ as $n\to \infty$, no?
$a^{r_m}$ remains bounded because the sequence $(r_m)$ is convergent so it is bounded.
Let $h:=\inf_{m\in\Bbb N} r_m$ and $L:=\sup_{m\in\Bbb N}r_m$, then $h\le r_m\le L$ for all $m\in\Bbb N$, and consequently $a^h\le a^{r_m}\le a^{L}$ (for $a>1$).
(Well, to complete the answer you need to show that a function defined by $x\mapsto a^x$ with $a>1$ is increasing.)