How do I prove that $$\lim_{(m,n) \to \infty} a_m^{b_n} = a^b$$
where $a,b \in \mathbb R$, $a_i,b_i \in \mathbb Q$, $a_m \to a$, $b_n \to b$ and $a$ and $b$ are not both zero, and $a_m >0$
I can prove it with $n$ or $m$ constant by using continuity, but not when both are limiting at the same time.
It depends on what you already "know".
First, there's another condition you probably meant to add in. You probably want to assume that $a_m > 0$. Otherwise, you can have things like $(-2)^{1/2}=\sqrt{2}i$ appear. If you assume that $a_m>0$ for all $m$, then $a_m\to a>0$ follows as well.
If you know continuity of the exponential and natural log functions, this is pretty straightforward.
$a_m^{b_n}=e^{\ln(a_m^{b_n})}=e^{b_n\ln(a_m)}$
Now assuming $e^x$ is continuous and $\ln(x)$ is continuous, we have that $y\ln(x)$ is continuous since the product of continuous functions is continuous and so $e^{y\ln(x)}$ is continuous since the composition of continuous functions is continuous.
Now that we know $f(x,y)=e^{y\ln(x)}$ is continuous, $(a_m,b_n) \to (a,b)$ implies that $f(a_m,b_n) \to f(a,b)$ and there's your result.