Step in a proof that $(a^x)^y = a^{xy}$. I'm trying to fill in the details in this proof, which seeks to show that for any real numbers $a, x, y$ with $a>0$ we have that $(a^x)^y = a^{xy}$. The sketch is as follows:
\begin{equation}
\begin{split}
(a^x)^y & = \sup \{ ( \sup\{ a^r : r∈\mathbb{Q}_{<x} \} )^s : s∈\mathbb{Q}_{<y} \}\\
& = \sup \{ \sup\{ (a^r)^s : r∈\mathbb{Q}_{<x} \} : s∈\mathbb{Q}_{<y} \}\\
& = \sup \{ (a^r)^s : r∈\mathbb{Q}_{<x} ∧ s∈\mathbb{Q}_{<y} \}\\
& = \sup \{ a^{t} : t∈\mathbb{Q}_{<xy}\}\\
& = a^{xy}\\
\end{split}
\end{equation}
I'm having trouble figuring out how the second equality (the transition from first to second line) is arrived at. In particular, we need to show that
$$(\sup \{a^r:r\in\mathbb{Q}_{<x}\})^s = \sup \{(a^r)^s:r\in\mathbb{Q}_{<x}\}.$$
I have been able to show that $(a^x)^s$ is indeed an upper bound of $\{(a^r)^s:r\in\mathbb{Q}_{<x}\}$, but I'm struggling in showing that it is its least upper bound. An idea is, for a fix $p$, to use the continuity of the functions $t\to t^p$ and $t\to p^t$, but I would like to avoid this since the benefit of this cumbersome approach to exponentiation is precisely that no Analysis beyond the basic properties of $\mathbb{R}$ is needed. I was wondering how else I could proceed.
 A: Note: The original problem correctly assumes $a>1$, so it suffices to prove the equality for $a>1$, $x,y> 0$. You may observe that ($\left(a^x\right)^y=\sup\{\sup\ldots\}$ does not even hold when $x<0$ due to $a^x<1)$. however, negative exponents can be reduced to positive case using for example $a^{-x}=\frac{1}{a^x}$.
You may use Bernoulli's inequality for rational exponent which has elementary proofs.
Then for every $\delta>0$ you may chose $r$ with $a^r>a^x-\delta$ (we'll impose additional restrictions on $\delta$ later).
Case 1: $s\ge 1$. We divide by $a^x$
$$
\frac{a^r}{a^x}>1-\frac{\delta}{a^x}
$$
Then useing Bernoulli:
$$\left(\frac{a^r}{a^x}\right)^s>\left(1-\frac{\delta}{a^x}\right)^s\ge1-\frac{s\delta}{a^x}$$
Finally:
$$\left(a^r\right)^s>\left(a^x\right)^s-\frac{s\left(a^x\right)^s\delta}{a^x}>\left(a^x\right)^s-\epsilon$$
The latest inequality can be proven for all $\epsilon>0$ as long as $\delta$ is carefully chosen, i.e. $1-\frac{\delta}{a^x}>0$ and $\epsilon>\frac{s\left(a^x\right)^s\delta}{a^x}$, conditions which can be met for $\delta$ "small enough".
Case 2: $0\le s<1$. We divide by $a^r$ and proceed similarly (with Bernoulli's inequality reversed).
Thus you said that you already proved that $\left(a^x\right)^s$ is an upper bound of $\{(a^r)^s:r\in\mathbb{Q}_{<x}\}$ and I proved that $\left(a^x\right)^s-\epsilon$ is not an upper bound, so $\left(a^x\right)^s$ is $\sup$
