Does anybody know a proof that the bases of two or more numbers with the same exponent can be multiplied together? (examples in the main text body). I.e. how can we simply assume that $5^{1/2}\times 6^{1/2} = 30^{1/2}$ (i.e. all we have done is multiplied the bases and applied the exponent to the answer).
In more general terms, how can we assume $a^e\times b^e = (ab)^e$? 
 A: For integer exponents, this is easily proved by induction in high school. For a more general (real) exponent, you start from the definition of exponent $e$:
$$a^e b^e\stackrel{\text{def}}{=}\exp(e\ln a)\exp(e\ln b)=\exp\bigl(e(\ln a+\ln b)\bigr)=\exp\bigl(e\ln(ab)\bigr)\stackrel{\text{def}}{=}(ab)^e.$$
A: Start with the specific case of positive integer $e$, and generalise from there, up to at least rational $e$ (and possibly beyond; but note that this property isn't true in general for complex $e$). For this case you have (by definition):
$$a^e = a \times a \times a ... \times a \,\,\, (e \text{ times}) $$
So you have:
$$(ab)^e = (ab) \times (ab) \times (ab) ... \times (ab) \,\,\, (e \text{ times})$$
and:
$$a^eb^e = a \times a \times a ... \times a \times b \times b ... \times b \,\,\, (e \text{ times of both } a \text{ and } b)$$
These are equal because multiplication is both associative and commutative, which together imply that we can reorder things as we please.
You can extend this to negative integer $e$ by treating $a^e$ as the fraction $\frac{1}{a^{-e}}$, and applying the above argument to the denominator. Similarly, it extends to fractional $e$ by taking it as a surd, and you can prove $\sqrt[n] a \sqrt[n] b = \sqrt[n]{ab}$ by raising both sides to the $n$, and again applying the positive integer case of $a^eb^e = (ab)^e$ to note that the results are equal, which means both sides must be equal since we define $\sqrt[n] a$ to be the positive root for such $n$ where that matters.
