How do we prove that the derivative of $x^r$ is $r x^{r-1}$ for all $r$?

$$\newcommand{\dd}{\frac{\mathrm{d}}{\mathrm{d}x}}$$

I haven't looked at your link, but it's important to note that the rule for taylor series and the rule you want to prove are not the same.

The first one says that if $r\in\mathbb R$, then $\dd x^r = rx^{r-1}$. For the taylor series rule, we only need that $r\in\mathbb N$, then $\dd x^r = rx^{r-1}$. This is much easier to prove (namely, we can use the difference quotient). Calculations with the difference quotient show that: \begin{align} \dd x^r & = \lim_{h\to 0}\frac{(x+h)^r-x^r}{h} \end{align} We have the binomial formula that $$(a+b)^n = \sum_{i = 0}^n \binom{n}{i}a^ib^{n-i}$$ We can use this to find that: \begin{align} \dd x^r & = \lim_{h\to 0}\frac{\sum_{ i = 0}^r\binom{r}{i}x^ih^{r-i}-x^r}{h} \\ &=\lim_{h\to 0}\frac{h\binom{r}{r-1}x^{r-1}+h^2\sum_{i = 0}^{r-2}\binom{r}{i}x^{i}h^{r-2-i}}{h} \\ &=\binom{r}{r-1}x^{r-1}=\frac{r!}{(r-1)!}x^{r-1} = rx^{r-1} \end{align}

As we used the binomial formula, we assume that $r\in\mathbb N$. In this way, it wasn't a strong enough proof to establish the result you want. It WAS strong enough to establish the power rule for differentiation taylor series. In this way, the proof isn't circular.

I see no such circular logic in the link you provided. The proof there only uses the Chain Rule and the derivative of $\ln(x)$ as established facts outside the scope of the proof. It reaches a point where it asks the reader to compute $$\frac{d}{dx}e^{r\ln(x)}$$ and then uses the Chain Rule, the derivative of $\ln(x)$, and the derivative of $e^x$ (established earlier in the proof) to calculate this derivative is $$r x^{r-1}\text{.}$$

The linked pdf (see comments to question by OP) gives a theory of exponential, logarithmic and general power functions of a real variable using the simplest possible approach (here simple means short and concise proofs and we assume that the theory of Riemann integrals is already developed and available for use) and this approach of defining logarithm via integral is very common in many textbooks of calculus/analysis.

There is no circularity involved in the proof of the derivative formula $$\frac{d}{dx}(x^{r}) = rx^{r - 1}\tag{1}$$ for $x > 0$ and $r \in \mathbb{R}$. It should however be noted that when $r$ is rational then the function $f(x) = x^{r}$ can be defined without any reference to $\log$ or $\exp$ and a proof of derivative formula then follows from algebraic inequalities. Moreover it is possible to extend this definition to define $x^{r}$ for any irrational $r$ and then the derivative formula $(1)$ also holds according to this extended definition. See this answer for more details on this approach which avoids logarithmic/exponential functions as far as the definition of $x^{r}$ and calculation of its derivative are concerned.