Closed form of $\sum_{k=1}^{n}\binom{n}{k} h^{(n-k)}(0)f^{(k-1)}(0)$ Is there a closed form for:
$$\sum_{k=1}^{n}\binom{n}{k} h^{(n-k)}(0)f^{(k-1)}(0)$$
where: $$h(x)=(1-x)^{\alpha}(A-Bx)^{\frac{1}{\gamma}-\alpha}$$ and $$f(x)=-x(1-x)^{-1-\alpha}(A-Bx)^{-(\frac{1}{\gamma}-\alpha)-1}$$ where $h^{(n-k)}(0)$ is the $(n-k)$th derivative of $h(x)$ at $x=0$.
I came across this trying to find the generating function for a distribution. I first solved a first order differential equation by integrating factor. As a result I obtained: 
$$y(x)=(1-x)^{\alpha}(A-Bx)^{\frac{1}{\gamma}-\alpha} \int -x(1-x)^{-1-\alpha}(A-Bx)^{-(\frac{1}{\gamma}-\alpha)-1}dx$$
 Next I need to find the Taylor series of $y(x)=\sum \frac{y^{(n)}(0)}{n!}x^n$. This requires the evaluation of the sum in this question. 
There seems to be a relation between $h$ and $f$ that makes me think(hope) there should be a nice closed form solution for it. 
I would greatly appreciate any suggestion!  
 A: Derivative of $h(x)$ seems to go something like
$$h^{(n)}(0)=(-1)^n(\alpha)_n A^{\frac{1}{\gamma}-\alpha}+(-B)^n\left({\frac{1}{\gamma}-\alpha}\right)_n A^{\frac{1}{\gamma}-\alpha-n}$$
And for $f(x)$ we have
$$f^{(n)}(0)=(-1)^n A^{-(\frac{1}{\gamma}-\alpha)-1}$$
So the sum will be
$$\sum_{k=1}^{n}\binom{n}{k}\left((-1)^{n-k}(\alpha)_{n-k} A^{\frac{1}{\gamma}-\alpha}+(-B)^{n-k}\left({\frac{1}{\gamma}-\alpha}\right)_{n-k} A^{\frac{1}{\gamma}-\alpha-n+k}\right)\left((-1)^{k-1} A^{-(\frac{1}{\gamma}-\alpha)-1}\right)\\
=\frac{(-1)^{n-1}}{A}\left(\sum_{k=1}^{n}\binom{n}{k}(\alpha)_{n-k}+\sum_{k=1}^{n}\binom{n}{k}\left({\frac{1}{\gamma}-\alpha}\right)_{n-k}\left(\frac{B}{A}\right)^{n-k}\right)$$
Exist a Sheffer sequence that relates binomial coefficient with a complementary pair of falling factorial but just for one falling factorial I dont found nothing. If you can relates binomial and a falling factorial in a closed form then maybe a closed form for the expression.
Maybe trying some finite differentiation.
UPDATE: well, wolfram-alpha says that exists a relation but Im not sure if this is a form more "closed" or not.
And for the second addend exists another relation.
