Binomial theorem proof for rational index without calculus

I have tried to find a proof of the binomial theorem for any power, but I am finding it difficult. One can obviously prove the integer index case using induction, but all of the approaches for ANY power seem to involve calculus usually the Maclaurin series. My point is that surely as the result $$\frac{d}{dx}(x^n)=nx^{n-1}$$ relies on the binomial theorem when taking the limit, this means that any attempt to prove the binomial theorem using calculus is circular. Is there a proof without calculus for ANY power?

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 $$ \frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}\tag{5} $$

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.