Let $\exp x=\sum_{n\geq 0} \frac {x^n}{n!}$. Let $e=\exp 1$.
Let $a,x\in \Bbb R$, $a>0$. We define $a^x=\sup \{a^r:r\in \Bbb Q, r<x\}$.
I've already proved that for $x=\frac pq \in \Bbb Q$, $a^x=\sqrt[q]{a^p}$ (where I've already proved the existence of roots, and $a^p=\prod_{i=1}^p a$), and using this, that $\exp (1)^r=\exp (r)$ for rational $r$.
Now want to show that $e^x$, i.e $\exp (1)^x=\exp (x)$, but the $\sup$ definition seems too hard to use.
Could anyone give me some hints?
Since $a = \exp(1) > 1$ we can prove the equality using your definition of $a^x$ as follows.
Let $(r_n)$ be an increasing sequence of rational numbers converging to $x$. You need to show that $a^{r_n} \to a^x$.
Once you have proved this then by continuity of $\exp(\cdot)$ we have
$$\exp(1)^x = \lim_{n \to \infty}\exp(1)^{r_n} = \lim_{n \to \infty}\exp(r_n) = \exp(\lim_{n \to \infty}r_n) = \exp(x).$$
With $a = \exp(1) > 1$ we see that $a^{r_n}$ is bounded above by $a^x$since
$$a^{r_n} \leqslant \sup_{n \in \mathbb{N}}a^{r_n} \leqslant \sup \{a^r:r\in \Bbb Q, r<x\} = a^x.$$
For rationals, $a^r$ is increasing. If $r > s$, then $a^r = a^{r-s}a^s$. Since $r - s > 0$ it is straightforward to show for $a > 1$ that $a^{r - s} > 1$ and, therefore $a^{r} > a^{s}.$ Hence, $(a^{r_n})$ is an increasing sequence.
We claim that $\sup_{n \in \mathbb{N}}a^{r_n} = a^x.$ Assume to the contrary that $\sup_{n \in \mathbb{N}}a^{r_n} < a^x.$ Then there exists a rational number $r < x$ such that $a^{r_n} < a^{r}$ for all $n$. However, since $r_n \to x$ there exists $m$ such that $r_m >r$. This implies $a^r < a^{r_m}$, a contradiction.
Therefore,
$$\sup_{n \in \mathbb{N}}a^{r_n} = a^x.$$
By monotone convergence
$$a^{r_n} \to a^x.$$