I have the term $$p=\left(\frac{1}{n-b}\right)^a\cdot[n]_a, 0<b<a,\ b \in \mathbb{R} \text{ fixed}$$ and want to find a tight lower bound such that I will then be able to solve for n.
For this, I am looking for a lower bound on the falling factorial $[n]_a=\frac{n!}{(n-a)!}$, given $n>a>1$, which should be tighter than $(n-a)^n$.
I used Stirlings Formula which got me
$$[n]_a \geq\left(\frac{n}{n-a}\right)^{n-a+0.5}\cdot\left(\frac{n}{e}\right)^a \exp\left(\frac{-12a-1}{12(12z^2-13n+z)}\right)$$
but it seems to me that the only useful term which would allow me to solve for $n$ in $p$ is $(\frac{n}{e})^a$, which is a lower bound more loose than $(n-a)^n$ for $n\gg a$.
Thanks in advance, Gitta