How to prove that $(a^n)^m = a^{nm}$ for non-integer exponents? This is a question that has been twinging me for a while. I have been told that $\sqrt{a} = a^{1\over{2}}, a\in\mathbb{R_+}$ and ${1\over{a}} = a^{-a}, a\in\mathbb{R_*}$ were nothing more that notations, not actual properties, and that, therefore, it could not / did not have to be proved or demonstrated.
The point I don't understand is that this "notation" can be used to retrieve, for instance, the formula of the derivative and primitives of the functions $x \longrightarrow \sqrt{x}$ and $x \longrightarrow {1\over{x}}$, using the power rule (${d\over{dx}}\sqrt{a} = {1\over{2\sqrt{a}}} = {1\over{2}}a\uparrow{-1\over{2}}$)
So my questions are whether $\sqrt{a} = a^{1\over{2}}$ is a real, proven "property" ; and if so, how it can be proved.
 A: This is a nice question.

*

*$ a^n $ is defined recursively for $ n \in \mathbb{N} = \mathbb{Z}_{\geq 0} $ for $ a $ real (or complex), with $ a^0 = 1 $ serving as a "base case" for the recursion.


*If we wish for exponentiation to be have the property $ a^{n+m} = a^na^m $ we are forced to define $ a^{-n} = 1/a^n $.  (Indeed let $ m = -n $.)  Here we are letting the desired properties drive a definition.  We are defining exponentiation for negative whole numbers so that exponentiation has some desirable properties.
After this point, there are two conceptually different ways we can go.


*[first way] Since $ \sqrt{a} \cdot \sqrt{a} = a $, it is consistent with the above exponentiation rules for non-negative real $ a $ to define $ \sqrt{a} = a^{1/2} $.  Similar reasoning forces us to define $ a ^ {p/q} $ = $\sqrt[q]{a^p} $ if we want exponentiation to continue to have the property $ a^{r + s} = a^r a^s $ for rational $ r $ and $ s $.


*[first way] Now that we have been essentially forced in how we define exponentiation for nonnegative real base $ a $ and exponent $ p/q $, we fill in the gaps for irrational real exponents $ t $ by insisting that $ a^t $ be a continuous function of $ t $.  This is the first time where we assert a new property we want exponentiation to have: we want it to be continuous.  Note that we are invoking the completeness of the reals to assert that $ p^{\mathrm{th}} $ roots exist.
There is another way to go about defining exponentiation beyond the integers.  Note that it relies on the fact that we've already defined it for the integers, so that we can define an infinite series


*[second way] Define the exponentiation function
$$
e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}}
$$
It satisfies $ e^0 = 1 $.  After some work, one can verify that it also happens to satisfy $ e^{x+y} = e^{x}e^{y} $ (!). Let $ \log(x) $ denote its inverse.  We then define
$$
a^r = e^{r \log(a)}
$$
and we have arrived immediately at exponentiation for all real exponents and non-negative base $ a $.  If you're willing to do the work, you can verify that it satisfies all the properties of exponents you've been taught.

Addendum (edit): You also asked about the rules of differential calculus.  You can prove them directly by implicit differentiation.  For example, if $ y = \sqrt{x} $ then $ y^2 = x $.  Differentiating both sides, we find that $ 2 y y' = 1 $, hence $ y' = 1 / (2y) = 1 / (2\sqrt{x}) $.  So you do not need the fractional exponent notation to prove facts about derivatives of root expreessions $ \sqrt[q]{x^p} $.  They can be proven directly by implicit differentiation by writing $ y = \sqrt[q]{x^p} $ and then rewriting $ y^q = x^p $ and implicitly differentiating.  If your calculus teacher failed to mention this, they did you a great disservice.
