How can I prove $\frac {d}{dx} {x^n} = n x^{n-1}$ for $ n \in \Bbb R$ without circular reasoning?

As stated in the comments we can use the fact that $$x^n=e^{n\ln{x}}$$ For all $x \in \mathbb{C}$ except $0$, $n \in \mathbb{C}$. Then, by using the chain rule, we have $$\frac{d}{dx}\Big(e^{f(x)}\Big)=f'(x)\cdot e^{f(x)}$$

So, $$\frac{d}{dx}\Big(x^n\Big)=\frac{d}{dx}\Big(e^{n\ln{x}}\Big)=\frac{n}{x}\cdot e^{n\ln{x}}=\frac{n}{x} \cdot x^n = n \cdot x^{n-1}$$

For $\boldsymbol{n\ge1}$

Bernoulli's Inequality, which is proven for integer exponents in this answer and extended to rational exponents in this answer using induction, says that for $n\ge1$, $$ 1+nx\le(1+x)^n\tag1 $$ From $(1)$, we get $$ 1+x\le\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag2 $$ Therefore, $$ x^n\left(1+\frac{nh}x\right)\le\overbrace{x^n\left(1+\frac hx\right)^n}^{(x+h)^n}\le x^ne^{nh/x}\tag3 $$ where the left inequality is $(1)$ and the right inequality is the $n^\text{th}$ power of $(2)$.

Subtracting $x^n$ and dividing by $h$ gives $$ x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac{e^{nh/x}-1}{nh/x}\tag4 $$ Applying $(9)$ gives $$ x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac1{1-nh/x}\tag5 $$ Then the Squeeze Theorem yields $$ \bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}}\tag6 $$

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Extending to $\boldsymbol{n\lt1}$

For smaller $n$, we can use the product rule and induction. That is, suppose we know that $(6)$ holds for some $n$, then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^{n-1} &=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1xx^n\right)\\ &=\frac1xnx^{n-1}-\frac1{x^2}x^n\\ &=(n-1)x^{n-2}\tag7 \end{align} $$ thus, $(6)$ holds for $n-1$.

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Bounds on $\boldsymbol{\frac{e^x-1}x}$

Taking $(2)$, substituting $x\mapsto-x$, and taking reciprocals yields that for $x\lt1$, $$ e^x\le\frac1{1-x}\tag8 $$ Combining $(2)$ and $(8)$, subtracting $1$ and dividing by $x$ gives $$ 1\stackrel{x\gtrless0}\lesseqgtr\frac{e^x-1}x\stackrel{x\gtrless0}\lesseqgtr\frac{1}{1-x}\tag9 $$