how can I prove that $$\sum_{n=0}^\infty{\frac{m_n\left(x;b,c\right)}{n!}t^n}=\left(1-t\right)^{-x-b}\left(1-\frac{t}{c}\right)^x$$
I tried using $$m_n\left(x;b,c\right)=\sum_{k=0}^{n}{\frac{n!}{(n-k)!\cdot k!}\left(x+b\right)_{n-k}\left(x-k+1\right)_{k}\left(-c\right)^{-k}}$$
this has created more confusing expression. Please, suggest a way to do this.
Thank you
Simplifying we obtain
$$m_n(x; b, c) = n! \sum_{k=0}^n {x+b+n-k-1\choose n-k} (-1)^k {x\choose k} c^{-k} \\ = n! \sum_{k=0}^n [t^{n-k}] \frac{1}{(1-t)^{x+b}} [t^k] \left(1-\frac{t}{c}\right)^x \\ = n! [t^n] \frac{1}{(1-t)^{x+b}} \left(1-\frac{t}{c}\right)^x.$$
This is the claim.