Convergence of binomial series I know that $f(x)=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$ converges for $|x|<1$
What I then have to show is that $(1+x)f'(x)=\alpha f(x)$ for $|x|<1$ and that any such $f$ is of the form $c(1+x)^\alpha$ for some constant c, and to use that fact to establish the binomial series.
I tried taking $$(1+x)f'(x)=(1+x)\left ( \sum_{n=0}^{\infty}\binom{\alpha}{n} x^n \right )'=(1+x)\left ( \sum_{n=0}^{\infty}\frac{\alpha !}{n!(\alpha-n)!} x^n \right )' = (1+x) \sum_{n=0}^{\infty}\frac{n \alpha !}{n!(\alpha-n)!} x^{n-1}=\alpha\sum_{n=0}^{\infty}\frac{ (\alpha-1) !}{(n-1)!((\alpha-1)-(n-1))!} (x^{n-1}+x^n)$$
But I'm not sure if my approach was right, and I'm not sure how to deal with the 2 $x$ terms if it was right.
I have no idea where to start for the second part.

$$\begin{align*}(1+x)f'(x)
&= (1+x) \sum_{n=1}^\infty \binom a n n x^{n-1}
= (1+x) \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n \\
&= \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n + \sum_{n=1}^\infty \binom a {n} n x^n \\
&= a + \sum_{n=1}^\infty \binom a {n+1} (n+1) x^n + \binom a {n} n x^n
= a + \sum_{n=1}^\infty a_n x^n
\end{align*}$$
where
$$\begin{align*}
a_n = \binom a {n+1} (n+1) + \binom a {n} n
&= \frac{a!}{(n+1)!(a-n-1)!}(n+1) + \frac{a!}{n!(a-n)!}n \\
&= \frac{a!a}{n!(a-n)!}
\end{align*}$$
Hence
$$(1+x)f'(x) = a + \sum_{n=1}^\infty a_n x^n = a + \sum_{n=1}^\infty \frac{a!a}{n!(a-n)!} x^n = af(x)
$$
The second part can be solved using logarithms, as RRL pointed out.
