Understand certain way of proving binomial series for $|x|<1$ and $\alpha \in \mathbb{Q}$

Question

I'm trying to understand a certain way of proving that

$$(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$$

for $\alpha \in \mathbb{Q}$ and $|x|<1$.

Now, my notes tackle that proof by saying that for $\alpha, \beta \in \mathbb{R}$:

$$\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n \cdot \sum_{n=0}^{\infty}\binom{\beta}{n}z^n = \sum_{n=0}^{\infty}c_nz^n$$

with $$c_n =\sum_{k=0}^{\infty}\binom{\alpha}{k}\binom{\beta}{n-k}=\binom{\alpha + \beta}{n} $$

and proceeding by proving that by induction.

Now, I can't even understand why proving the second notion would prove the initial theorem, so if anybody could explain or provide further material since I can't find anything about that particular way of proof, I'd be very thankful.

Answer 1

My guess is that what you wrote on the second paragraph is not meant to be a proof of the fact that $$(1+x)^\alpha=\sum_{n=0}^{\infty}\binom\alpha nx^n,\tag1$$but a proof of the fact that, if you define $(1+x)^\alpha$ by $(1)$, then$$(1+x)^\alpha(1+x)^\beta=(1+x)^{\alpha+\beta},\tag2$$since what it says there is precisely what is needed to prove that $(2)$ holds.

Answer 2

Here's an alternate proof, if you're interested.

One may show that $$p_n(\alpha)=\left[\frac{\partial}{\partial x}x^\alpha\right]_{x=1}=\prod_{r=1}^{n}(\alpha-r+1).$$ So then the Taylor series for $x^\alpha$ about $x=1$ is given by $$x^\alpha=\sum_{k\ge0}\frac{p_k(\alpha)}{k!}(x-1)^k.$$ We notice that $${n\choose k}=\prod_{i=1}^n\frac{n+1-i}{i}.$$ So we have $$(1+x)^\alpha=\sum_{k\ge0}{\alpha\choose k}x^k.$$