# Finite sum of the Pochhammer symbols (falling factorials)

Solving a problem I stuck with a sum $$\sum_{k=1}^{n}k \, a^\underline{k}.$$ With a finite difference trick $\Delta_k(a^\underline{k}) = a^\underline{k} \, (a - k - 1)$, written in a book "Concrete Mathematics" I could simplify it to $$\sum_{k=1}^{n}a^\underline{k}.$$ But this can be written as the finite sum of the inverse factorials $$\sum_{k=1}^{n}a^\underline{k} = a! \, \sum_{k=1}^{n} \frac{1}{(a-k)!} = a! \, \sum_{k=a-1}^{a-n} \frac{1}{k!}.$$ My colleague told me that this sum can not be written in a closed form. So I am asking if there is such a form. If not, may be there is another way? The approach with summing by parts gives bad result because of the form of the finite difference of the function.

P. S. Also there is an interesting fact, when comparing to a continuous analogue of this function: $D_x(a^x) \sim c \, a^x$, but in discrete case it is $D_x(a^x) \sim c \, x \, a^x$.

$$\sum_{k=1}^{n}a^\underline{k} = a! \, \sum_{k=1}^{n} \frac{1}{(a-k)!} =-1+e \Gamma (a+1,1)-e a\frac{ \Gamma (a)\, \Gamma (a-n,1)}{\Gamma (a-n)}$$ where appear the complete and incomplete gamma functions.