Here's the definition provided in Richard Silverman's Modern Calculus and Analytic Geometry:
Suppose that $x \in \mathbb{R}$ and $a > 0$. Let $\{r_n\}$ be a sequence of rationals such that:
$$\lim_{n \to \infty} r_n = x$$
Then, we define the exponential of base $a$ as follows:
$$a^x = \lim_{n \to \infty} a^{r_n}$$
I've already demonstrated that this limit does actually converge so that's not really the issue here. This definition makes sense. Now, I'm trying to prove that the following result holds:
Let $x_1,x_2 \in \mathbb{R}$. Suppose that $a > 0$. Then:
$$(a^{x_1})^{x_2} = a^{x_1x_2}$$
I have something of a proof attempt but it looks dubious. Let me just show it below:
Proof Attempt:
Let $\{r_n\}$ and $\{s_n\}$ be sequences of rationals such that:
$$\lim_{n \to \infty} r_n = x_1$$
$$\lim_{n \to \infty} s_n = x_2$$
Then, by definition, we have:
$$a^{x_1} = \lim_{n \to \infty} a^{r_n}$$
Now, we have:
$$(a^{x_1})^{x_2} = \lim_{n \to \infty}[(\lim_{n \to \infty} a^{r_n})^{s_n}]$$
We cannot simplify this to one limit because the two limits do not converge at the same time. That is, for any fixed $\epsilon > 0$:
$$\exists N_1 > 0 : n > N_1 \implies |a^{r_n}-a^{x_1}| < \epsilon$$
$$\exists N_2 > 0 : n > N_2 \implies |a^{s_n}-a^{x_2}| < \epsilon$$
where $N_1$ is not necessarily equal to $N_2$. Now, we take our fixed $\epsilon$ and define $N = \max(N_1,N_2)$. Then, for any $n > N$, the two inequalities above hold and the sequences converge at the same time. So, it follows that we can simplify the limit expression we have above so that:
$$(a^{x_1})^{x_2} = \lim_{n \to \infty}[(a^{r_n})^{s_n}] = \lim_{n \to \infty}(a^{r_ns_n}) = a^{x_1x_2}$$
That proves the desired result.
I think this is sort of-kind of a proof but not really. I'm really uncomfortable about the simplification I made and, intuitively, it makes sense but I'm not sure if this is rigorous at all.
Essentially, I feel like there's probably more to this that I'm not really seeing and I hope someone can tell me if I'm right to feel that way about the argument above.
If you want to write a proof following the same line as the argument above, then you are welcome to but please put it in spoilers and give hints for me to fix my argument above instead.
Edit:
I'm going to try showing that for any $x \in \mathbb{R}$ and any $r \in \mathbb{Q}$, it is the case that:
$$(a^x)^r = a^{xr}$$
where $a > 0$. Since $r \in \mathbb{Q}$, it follows that $r = \frac{p}{q}$, where both $p$ and $q$ are integers and $q > 0$. With that in mind:
$$(a^x)^r = (a^x)^{\frac{p}{q}} = [(a^x)^p]^{\frac{1}{q}}$$
$$(a^x)^r = [a^{px}]^{\frac{1}{q}}$$
That holds because $(a^x)^n = a^{nx}$ for any integer $n$> This can be easily shown using induction from the fact that $a^{x_1+x_2} = a^{x_1} \cdot a^{x_2}$ for any real numbers $x_1$ and $x_2$.
$$(a^x)^r = [a^{\frac{pxq}{q}}]^{\frac{1}{q}} = [a^{\frac{px}{q}}]^{\frac{q}{q}} = a^{\frac{px}{q}} = a^{rx}$$
So, with this, we have:
$$(a^{x_1})^{x_2} = \lim_{n \to \infty} (a^{x_1})^{s_n} = \lim_{n \to \infty} a^{x_1s_n} = a^{x_1x_2}$$
since $x_1s_n$ converges to $x_1x_2$ as $n \to \infty$. This proves the desired result.