Binomial Theorem/General Binomial Theorem/Proof 1

Proof
Let $R$ be the radius of convergence of the power series:
 * $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {n!} x^n$

By Radius of Convergence from Limit of Sequence:


 * $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \frac {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n}\right)}\right|} {\left({n + 1}\right)!} \frac {n!} {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n + 1}\right)}\right|}$

Thus for $\left|{x}\right| < 1$, Power Series Differentiable on Interval of Convergence applies:


 * $\displaystyle D_x f \left({x}\right) = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {n!} n x^{n - 1}$

This leads to:

Gathering up:
 * $\left({1 + x}\right) D_x f \left({x}\right) = \alpha f \left({x}\right)$

Thus:
 * $D_x \left({\dfrac {f \left({x}\right)} {\left({1 + x}\right)^\alpha}}\right) = -\alpha \left({1 + x}\right)^{-\alpha - 1} f \left({x}\right) + \left({1 + x}\right)^{-\alpha} D_x f \left({x}\right) = 0$

So $f \left({x}\right) = c \left({1 + x}\right)^\alpha$ when $\left|{x}\right| < 1$ for some constant $c$.

But $f \left({0}\right) = 1$ and hence $c = 1$.