How to define an irrational to the power an irrational?

First, show there's an $n$th root by considering the l.u.b. of numbers whose $n$th powers are no more the target number.
Then, show that the Then, show that the nth root is unique, by noticing that nth power is one-to-one function.th root is unique, by noticing that Then, show that the nth root is unique, by noticing that nth power is one-to-one function.th power is one-to-one function.
Define, temporarily,
$x^{\frac{k}{l}}=(x^{\frac{1}{l}}{)}^k$
Then, notice that for any $x$, and integer exponents, $x^{ab}=(x^a{)}^b=(x^b{)}^a$, because this kind of exponentiation is defined by multiplication.
Therefore if $r$ is rational, with two different representations $r=\frac{p}{q}=\frac{m}{n}$, then
$\begin{array}{rcl}(x^r{)}^{qn}& =& (x^{\frac{p}{q}}{)}^{qn}=(x^{\frac{1}{q}}{)}^{qpn}=[(x^{\frac{1}{q}}{)}^q{]}^{pn}=x^{pn}\end{array}$
But also
$\begin{array}{rcl}(x^r{)}^{qn}& =& (x^{\frac{m}{n}}{)}^{nq}=(x^{\frac{1}{n}}{)}^{nmq}=[(x^{\frac{1}{n}}{)}^n{]}^{mq}=x^{mq}\end{array}$
Now, $np=mq$, since the fractions are equal, so both ways of expressing $r$ as a fraction give the same value for of $x^r$ (again because raising to integer exponents is one-to-one). Since there's a well-defined rational exponent, all you have to do is define:
$a^x=sup\{a^r:r\in \mathbb{Q},r\le x\}$