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**arbolis** Prove that the function $\displaystyle f(z)=1+ \sum _{n=1}^{\infty} \frac{a(a-1)...(a-n+1)}{n!}z^n $ is analytic when $\displaystyle |z|<1$ and that its derivative is $\displaystyle \frac{a f(z)}{1+z}$. Then demonstrate that the derivative of $\displaystyle (1+z)^{-a} f(z)$ is worth $\displaystyle 0$ and deduce that $\displaystyle f(z)=(1+z)^a$. Note that $\displaystyle (1+z^a)=exp(a Log(1+z))$.

My attempt: $\displaystyle f(z)= \sum _{n=1}^{\infty} \frac{(a-0)(a-1)(a-2)...[a-(n-1)]}{n!} z^n=az+\frac{a(a-1 z^2)}{2}+\frac{a(a-1)(a-2) z^3}{3}+...=\alpha z + \beta z^2 + \gamma z ^3 +...$ where $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$,... $\displaystyle \in \mathbb{R}$. So it seems to me that f is analytic for any value of z though I don't know how to formally prove it. So I don't understand why they bother with |z|<1.

I'd like to tackle this exercise and eventually finish it with your help. Thanks for any comment/tip.