Binomial theorem proof for rational index without calculus
You can deduce the series expansion of $(1+x)^{1/2}$ without calculus.
Suppose $(1+x)^{1/2} = \sum_{r=0}^\infty a_rx^r$. Squaring both sides gives
$1 + x = a_0^2 + 2a_0a_1x + (a_1^2 + 2a_2a_0)x^2+\cdots$
Now equating coefficients of $x^r$ gives:
$a_0^2=1$ (and we choose $a_0=+1$ to get the positive branch of $(1+x)^{1/2}$)
$2a_0a_1 = 1$, so $a_1 = \frac12$
$a_1^2 + 2a_2a_0 = 0$, so $a_2 = -\dfrac{a_1^2}{2a_0} = -\frac18$
and so on.
In theory, you can use the same technique to deduce the series expansion of $(a+x)^{1/q}$ for any positive integer $q$. And from there you can raise the series to the $p$th power to get $(1+x)^{p/q}$ for any integers $p,q$. But it rapidly becomes unmanageable as $q$ gets larger.
Suppose $\alpha\ge1$. Using Bernoulli's Inequality (which can be proven by induction for integer exponents, and easily extended for rational exponents, then extended by continuity for real exponents), we have for $|h|$ small enough so that $\frac{\alpha|h|}{x}\lt1$, $$ \begin{align} \frac{(x+h)^\alpha-x^\alpha}{h} &=x^\alpha\frac{\left(1+\frac{h}{x}\right)^\alpha-1}{h}\\ &\ge x^\alpha\frac{\left(1+\frac{\alpha h}{x}\right)-1}{h}\\[12pt] &=\alpha x^{\alpha-1}\tag{1} \end{align} $$ Furthermore, for $|h|$ small enough so that $0\lt\frac{\alpha|h|}{x-|h|}\lt1$, $$ \begin{align} \frac{(x+h)^\alpha-x^\alpha}{h} &=x^\alpha\frac{\left(1+\frac{h}{x}\right)^\alpha-1}{h}\\ &=x^\alpha\frac{\frac1{\left(1-\frac{h}{x+h}\right)^\alpha}-1}{h}\\ &\le x^\alpha\frac{\frac1{\left(1-\frac{\alpha h}{x+h}\right)}-1}{h}\\[9pt] &=\alpha x^{\alpha-1}\frac{x}{x-(\alpha-1)h}\tag{2} \end{align} $$ Thus, using the Squeeze Theorem with $(1)$ and $(2)$, we have $$ \frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\lim_{h\to0}\frac{(x+h)^\alpha-x^\alpha}{h}=\alpha x^{\alpha-1}\tag{3} $$
For $\alpha\lt1$, we have from $(3)$ that $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$. Then, because $2-\alpha\gt1$, the product rule says $$ \begin{align} 2x &=\frac{\mathrm{d}}{\mathrm{d}x}x^{(2-\alpha)+\alpha}\\ &=x^{2-\alpha}\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha+(2-\alpha)x^{1-\alpha}x^\alpha\tag{4} \end{align} $$ Finally, $(4)$ and algebra say that $$ \frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}\tag{5} $$
The fact that $\frac{d}{dx} x^n = nx^{n-1}$ for any $n \in \mathbb{R}$ does not rely on the binomial theorem. One can instead use the chain rule as follows:
Consider that $$\frac{d}{dx} \ln\left(x^n\right) = \frac{1}{x^n}\frac{d}{dx}x^n$$
by the Chain Rule. However,
$$\frac{d}{dx} \ln(x^n) = n \frac{d}{dx} \ln(x) = \frac{n}{x}$$ Thus $\frac{d}{dx} x^n = \frac{n}{x} x^n = nx^{n-1}$.
I know that this does not address the problem of proving the binomial theorem, but hopefully this helps with questions of possible circularity.