Generalised Binomial Theorem Intuition It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:
$$
(x+y)^n=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r
$$
Unfortunately, I wasn't able to find any clear explanation of how to get from the regular theorem to the generalised one, the only proof I found being based on some obscure mathematics, while my math book entirely skips the explanation. 
Hence, my question is how do you prove the generalised theorem by deriving it from the regular theorem or otherwise?

Here is one straightforward approach, mentioned in Lang's Analysis book.
Let $ \alpha \in \mathbb{R} $ and $ x \in (-1,1) $.
We'll try to show $$ (1+x)^\alpha = 1 + \binom{\alpha}{1}x + \binom{\alpha}{2} x^2 + \ldots $$ holds (where for $ k \in \mathbb{Z}_{>0} $, $ \binom{\alpha}{k} := \frac{\alpha (\alpha - 1) \ldots (\alpha - (k-1))}{k!} $)
Writing $ (1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2} x^2 + \ldots + \binom{\alpha}{n-1} x^{n-1} + R_n (x) $, we now need to show $ R_n (x) \to 0 $ as $ n \to \infty $.

Recall integral version of Taylor's theorem

Let $ f : I \to \mathbb{R} $ be a $ C^n $ function on open interval $ I $, and say $ a, b \in I $. Then $$ f(b) = f(a) + f'(a) (b-a) + \ldots + \dfrac{f^{(n-1)}(a)}{(n-1)!} (b-a)^{n-1} + \int_{a}^{b} \dfrac{(b-t)^{n-1}}{(n-1)!} f^{(n)}(t) dt $$

So applying it to $ f : (-1, \infty) \to \mathbb{R} $ given by $ f(t) = (1+t)^\alpha $ and with $ a = 0 $,
$$ \begin{align} R_n (x) &= \int_{0}^{x} \dfrac{(x-t)^{n-1}}{(n-1)!} \alpha (\alpha - 1) \ldots (\alpha -(n-1)) (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \end{align} $$

Estimating $\binom{\alpha}{n} $ term :
If $ \alpha = 0 $, $ \binom{\alpha}{n} $ is trivially $ 0 $. So lets take $ \alpha \neq 0 $.
$\begin{align} \left|\binom{\alpha}{n}\right| &= \left|\dfrac{\alpha (\alpha-1) \ldots (\alpha - (n-1))}{n!} \right| \\ &= \left| \dfrac{1}{n} \alpha \left( \dfrac{\alpha}{1} - 1 \right) \left( \dfrac{\alpha}{2} - 1 \right) \ldots \left( \dfrac{\alpha}{n-1} - 1 \right) \right| \\ &\leq \dfrac{1}{n} |\alpha| \left( \dfrac{|\alpha|}{1} + 1 \right) \left( \dfrac{|\alpha|}{2} + 1 \right) \ldots \left( \dfrac{|\alpha|}{n-1} + 1 \right) \\ \end{align}$
Since each factor in RHS is $ > 0 $, we can write it as
$$ e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] }. $$
The exponent here is
$\begin{align} &- \ln n + \ln|\alpha| + \ln\left(\frac{|\alpha|}{1} + 1\right) + \ln\left( \frac{|\alpha|}{2} + 1 \right) +   \ldots + \ln\left( \frac{|\alpha|}{n-1} + 1 \right) \\ &\leq - \ln n + \ln |\alpha | + \frac{|\alpha|}{1} + \frac{|\alpha|}{2} + \ldots + \frac{|\alpha|}{n-1} \end{align} $
[because $ \ln (1+x) \leq x $ for $ x \geq 0 $. This in turn is because $ x - \ln(1+x) $ is $ 0 $ at $ x = 0 $, and its derivative $ 1 - \frac{1}{1+x} $ is $ > 0 $ on $ (0, \infty) $]
$\begin{align} &\leq -\ln n + \ln |\alpha| + |\alpha| \left( 1 + \ln (n-1) \right) \end{align} $
[because $ 1 + \frac{1}{2} + \ldots + \frac{1}{n-1} $ $= 1 + \int_{1}^{2} \frac{1}{2} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{n-1} dt $ $ \leq 1 + \int_{1}^{2} \frac{1}{t} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{t} dt $ $ = 1 + \ln (n-1) $. This is also clear visually].
So finally,
$\begin{align} \left| \binom{\alpha}{n} \right| &\leq e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] } \\ &\leq e^{-\ln n + \ln |\alpha| + |\alpha| (1 + \ln (n-1) ) } \\ &= \frac{1}{n} |\alpha| e^{|\alpha|} (n-1)^{|\alpha|} ( \leq |\alpha| e^{|\alpha|} n^{|\alpha| - 1} ) \end{align} $
Especially,
$$ \fbox{$\left|\binom{\alpha}{n}\right| = \mathcal{O} (n^{|\alpha|-1}) $ } $$

Estimating $R_n(x)$ :
$ \underline{\text{CASE-1}} \, (x \in [0,1)) $
In the integral as $ t $ varies from $ 0 $ to $ x $, $ (1+t)^{\alpha - n} \leq (1+t)^{\alpha} \leq 2^\alpha $, so
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n}\right| \int_{0}^{x} (x-t)^{n-1}(1+t)^{\alpha - n} dt \\ &\leq n \left| \binom{\alpha}{n} \right| 2^\alpha \int_{0}^{x} (x-t)^{n-1} dt \\ &= \left| \binom{\alpha}{n} \right| 2^\alpha x^n   \end{align}$
Now RHS is $ \leq K n^{|\alpha|-1} x^n $, and $ n^{|\alpha|-1} x^n $ goes to $ 0 $ as $ n \to \infty $ [if $ x = 0 $, trivial. Else write $ x $ as $ \frac{1}{1+y} $ with $ y > 0 $ and expand the denominator], as needed.
$\underline{\text{CASE-2}} \, (x \in (-1, 0)) $
$\begin{align} R_n (x) &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{-|x|} (-|x| - t)^{n-1} (1+t)^{\alpha - n} dt \end{align}$
Substituting $ s = (-t) $, the integral becomes
$\begin{align} R_n(x) &= n \binom{\alpha}{n} \int_{0}^{|x|} (-|x| + s)^{n-1} (1-s)^{\alpha - n} (-ds) \\ &= n \binom{\alpha}{n} (-1)^n \int_{0}^{|x|} (|x|-s)^{n-1} (1-s)^{\alpha - n} ds \end{align}$
and now keeping in mind $ |x| \in (0,1) $ and that $ s $ varies from $ 0 $ to $ |x| $ in the integral,
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} \left(\dfrac{|x|-s}{1-s} \right)^{n-1} (1-s)^{\alpha - 1} ds \\ &\leq n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} |x|^{n-1} (1-s)^{\alpha -1}ds \\ &\leq L n^{|\alpha|} |x|^{n-1}\end{align}$
and $ n^{|\alpha|} |x|^{n-1} $ goes to $ 0 $ as $ n \to \infty $, as needed.
