Proving power rule for $x^n$ with arbitrary positive $n>0$ 
How to prove that $\frac{d}{dx} x^n = nx^{n-1}$ for every $n>0$ (possibly fractional)?

Context
It was already shown that $\frac{d}{dx} x^n = nx^{n-1}$ for positive integer $n$.  My friend told me that the general case is pretty tricky because in order to prove it, I firstly need to be able to define $x^n$ in each case (rational and irrational real numbers). He made an example using the exponential function,  $$e^x = 1 + x + \frac12 x^2 + \frac16 x^3+\dots $$ However, I have not yet covered Taylor series or Maclaurin series.

We want to find the derivative of $x^a$, for positive $x$.  Note that $x^a=\exp(a\ln x)$. (For general $a$, that is precisely how $x^a$ is ordinarily defined.).
Differentiate, using the Chain Rule. 
We get $\frac{a}{x}\exp(a\ln x)$. This is $\frac{a}{x}x^a$, which is $ax^{a-1}$.  That is the rule that we already knew held when $a$ is a positive integer.
