Proof of the derivative of $x^n$

It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.

I'd rewrite it using this:

For $n\ge 2$ there exists some polynomial $P(x,h)$ such that $$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$

The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x \rightarrow y'=e^x$.

From the inverse function differentiation rule we find $y=\log x \rightarrow y'=\dfrac{1}{x}$ and (using the chain rule):

$$ y=x^a =e^{a \log x} \rightarrow y'=e^{a \log x} (a\log x)'=e^{a \log x} \dfrac{a}{x}=ax^{a-1} $$