Question. Prove $(1+x)^r=1+rx+\frac{r(r-1)}{2!}x^2+\frac{r(r-1)(r-2)}{3!}x^3+\cdots$, where $r$ is a real number, without using Taylor (including Maclaurin) series.
Origin. In a class, the teacher was proving the derivative of $x^n$ where $n$ is a real number by the definition of differentiation $\lim_{h\to 0}\frac{\operatorname{f}(x+h)-\operatorname{f}(x)}{h}$ . During the process, the binomial theorem with real exponent was used. However, when I looked up the proof for the binomial theorem, every single one used the Taylor series which assumed the polynomial differentiability which was to be proven! By the way, anyone knows the technical term for such a situation, i.e. it is like using a corollary to prove the theorem from which the corollary is derived?
Attempt. About seven months ago one of my teachers presented a proof but I forgot how it went exactly. I only remember that it started with $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, and somehow he derived the series expansion of $\ln(x)$ from that, and then I think that he expressed $(1+x)^r$ as $e^{r\ln(1+x)}$. So the problem boils down to how to derive the series expansion of $\ln(x)$ from the series expansion of $e^x$.
The r.h.s can be written
$$I_r(x) = \sum_{k =0}^\infty A_{k,r} \frac{x^k}{k !},$$ where $A_{0, r} = 1, \,\, A_{k,r} := r (r-1)\cdots(r-k+1), k\geq 1.$
Working on the sum itself, we get $$I'_r(x) = \sum_{k =1}^\infty A_{k,r} \frac{x^{k-1}}{(k-1) !}.$$ Multiplying with $(1+x),$ we get $$(1+x) I'_r(x) =\sum_{k =1}^\infty A_{k,r} \frac{x^{k-1}}{(k-1) !} + \sum_{k =1}^\infty A_{k,r} \frac{x^k}{(k-1) !}.$$ Now, we change the index inside each sum, $$(1+x) I'_r(x) =\sum_{k =0}^\infty A_{k+1,r} \frac{x^k}{k !} + \sum_{k =0}^\infty k A_{k,r} \frac{x^k}{k !}.$$
The last step is to prove that $$A_{k+1, r} + k A_{k, r} = r A_{k , r}.$$ Finally, we get $$(1+ x) I'_r(x) = rI_r(x)$$ which is a first order DE that you can solve easily.