Generalized binomial theorem 
Prove that:
  $$(1+x)^{\alpha}=\sum_{n=0}^{+\infty}{\alpha \choose n} x^n$$
  for $x\in[0;1), \alpha \in\mathbb{R}$ based on Taylor's theorem with Lagrange remainder.

I don't feel such proofs. Isn't it kind of.. obvious? That when we expand $(1+x)^{\alpha}$ into series we get right side? I really want to understand this.
 A: You have that if 
$$y=(1+x)^\alpha$$
then
$$y'(0)=\alpha$$
$$y''(0)=\alpha(\alpha-1)$$
$$\cdots=\cdots$$
$$y^{(n)}(0)=\alpha(\alpha-1)\cdots(\alpha-n+1)$$
$$y^{(n)}(0)=\frac{\alpha!}{(\alpha-n)!}$$
You can prove this last generalization by induction on $n$. The last equation is written for the sake of "order", but if you're not comfortable with it, you can use the $\Gamma$ function to denote the factorials.
Then
$$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k \frac{\alpha!}{(\alpha-n)!}\frac{x^n}{n!}$$
Then using the binomial coefficient notation
$$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k {\alpha \choose n}x^n$$
What you need now is that the series converges for $|x|<1$. 
In my opinion you can argue as follows:
Fix $x$. Then the series
$$\lim\limits_{k \to \infty} \mathcal{T}_k(y,0)=\sum\limits_{n=0}^\infty {\alpha \choose n}x^n$$
will converge when 
$$ \lim\limits_{n \to \infty} \left|\frac{{\alpha \choose n+1}}{{\alpha \choose n}}\frac{x^{n+1}}{x^n}\right|<1$$
$$ \lim\limits_{n \to \infty} \left|\frac{{n+1}}{n-\alpha}x\right|<1$$
$$  \left|x\right|<1$$
ADD: Here's a part of an answer to a similar question:
Assume
$${\left( {1 + x} \right)^{\alpha}} = \sum\limits_{k = 0}^\infty  {{a_k}{x^k}} $$
Since if two series
$$\eqalign{
  & \sum {{a_k}{{\left( {x - a} \right)}^k}}   \cr 
  & \sum {{b_k}{{\left( {x - a} \right)}^k}}  \cr} $$
sum up to the same function then
$${a_k} = {b_k} = \frac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}$$
for every $k \leq 0$, we can assume:
$$a_k = \dfrac{f^{(k)}(0)}{k!}$$ 
Putting $y = {\left( {1 + x} \right)^{\alpha}}$ we get
$$y'(0) = \alpha$$
$$y''(0) = \alpha(\alpha-1)$$
$$y'''(0) = \alpha(\alpha-1)(\alpha-2)$$
$$y^{{IV}}(0) = \alpha(\alpha-1)(\alpha-2)(\alpha-3)$$
We can prove in general that
$$y^{(k)}= \alpha(\alpha-1)\cdots(\alpha-k+1)$$
or put in terms of factorials
$$y^{(k)}(0)= \frac{\alpha!}{(\alpha-k)!}$$
This makes
$$a_k =  \frac{\alpha!}{k!(\alpha-k)!}$$ 
which is what we wanted.
$${\left( {1 + x} \right)^\alpha } = \sum\limits_{k = 0}^\infty {\alpha \choose k}  {{x^k}} $$
You can prove this in a more rigorous manner by differential equations:


*

*Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k}  {{x^k}}$ and prove the radius of convergence is 1.

*Show that $f(x)$ is the solution to the ODE $$y' - \frac{\alpha }{{x + 1}}y = 0$$ with initial condition $f(0)=1$. 

*By the theorem that the solution to the linear equation


$$y'+P(x)y=R(x)$$
with initial conditions $f(a) = b$ is unique, you can prove the assertion. (prove that 
$y = {\left( {1 + x} \right)^{\alpha}}$ also satifies the equation and you're done.)
