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A teaching question. What is or might be simplest form for rational exponents? My search turns up little for the concept other than a translation of the simplest form for n-th root radicals. But that leads to ugly things such as...
$\large{5^\frac{7}{3}\rightarrow 5^2\centerdot 5^\frac{1}{3}}$
$\large{x^{-\frac{2}{3}}\rightarrow x^{-1}\centerdot x^\frac{1}{3}}$
...when the whole point of rational exponents is beauty and elegance.
What's "simplest" is highly context dependent - context which you have not supplied. However, one important aspect of simplicity is computational effectivity. In order to have any hope of effectively computing with such radicals it is essential to normalize them so that any algebraic relationships are made explicit. For example, suppose that you are computing with expressions involving $\:5\:$ raised to the powers $\ 1/6,\ 1/10,\:$ and $\:1/15\:.\:$ Then such expressions can all be expressed as polynomials in the single radical $\:5^{1/30}\:,\:$ e.g. $\rm\:5^{1/6} = (5^{1/30})^5\:.\:$ Said in the language of field theory $\rm\ \mathbb Q(5^{1/6},\:5^{1/10},\:5^{1/15}) \subset \mathbb Q(5^{1/30})\:.\:$ Therefore arithmetic operations on such radical expressions reduce to simple efficient polynomial operations, viz. operations in the ring $\rm\ Q[x]\ (mod\ x^{30}-5)\:.\:$
Such normalization is even more crucial when working with multiple radicands. For example $\ \sqrt{6}\ \sqrt{10}\ \sqrt{15}\ =\ 30\:,\:$ so this algebraic dependence needs to be eliminated by choosing two of the three radicands as a basis, say $\:\sqrt{6},\ \sqrt{10}\:,\ $ then eliminating $\:\sqrt{15}\ $ using $\rm\ \sqrt{15}\ =\ \sqrt{6}\ \sqrt{10}/2\:.\:$ Developing effective algorithms for computing with such radicands (e.g. denesting) is a highly nontrivial task, based on the Galois theory of radical extensions (Kummer theory). This is the theory at the basis of such effective algorithms implemented in computer algebra systems.
"Lowest terms" for a rational is usually understood to mean $a/b$, with $a$ and $b$ integers, $b\gt 0$, and $\gcd(a,b)=1$. (That is, the "sign" goes with the numerator).
For $p/q$ written in lowest terms, the definition of $a^{p/q}$ is that $a^{p/q} = r$ where $r$ is the unique nonnegative real number such that $r^q = a^p$; that is, $\displaystyle a^{p/q} = \sqrt[q]{a^p}$. This includes the cases with $p\gt q$ and with $-p\gt q$. If you want only fractional exponents of the form $p/q$ with $0\lt p\lt q$, then of course this can be done fairly easily, but it does not seem to me to lead to much simplification for algebraic manipulations (though it may for actual computations).
Rational exponents in lowest terms with even denominator are only, a priori, defined for positive bases when working in the reals.
