Proof of derivative of $x^n$ for all real n and real x

Question

I've seen the following proof for the derivative of $x^n$: $$ (x^n)' = (e^{ln(x)n})' = x^n \cdot \frac{n}{x} = nx^{n-1} $$ But the equality $x^n = e^{ln(x)n}$ is only valid for $x \in (0,\infty)$. How can one prove this property for all $x \in \mathbb{R}$ (Without using a complex logarithm)?

Answer 1

The claim isn't exactly true since x^n is not defined for two real numbers. For example: (-1)^0.5 is complex