Asymptotic behaviour of generalized binomial coefficient $\frac{an(an-1)...(an-n+1)}{n!}$ Let $a\in(0,1)$. What is the asymptotic behaviour of $\frac{an(an-1)...(an-n+1)}{n!}$ as $n\rightarrow\infty$?
It looks like it might be possible to express this in terms of gamma functions and use Stirling's approximation.
It is clear that $\frac{an(an-1)...(an-n+1)}{n!}=\frac{\Gamma(an+1)}{\Gamma(an-n+1)\Gamma(n+1)}$. But Stirling's approximation doesn't work for a gamma function with a negative argument.

You can use the reflection formula and the known asymptotics for the gamma function to obtain
\begin{align*}
\frac{{\Gamma (an + 1)}}{{\Gamma (an - n + 1)\Gamma (n + 1)}} & = \frac{{\Gamma (an + 1)}}{{\Gamma (1 - (1 - a)n)\Gamma (n + 1)}} \\ & = \frac{{\sin (\pi (1 - a)n)}}{\pi }\frac{{\Gamma (an + 1)\Gamma ((1 - a)n)}}{{\Gamma (n + 1)}}\\ &
\sim \sqrt {\frac{{2a}}{{\pi (1 - a)n}}} \sin (\pi (1 - a)n)((1 - a)^{1 - a} a^a )^n ,
\end{align*}
as $n\to +\infty$ with any fixed $0<a<1$.
