for $n,m\in \mathbb{N}$ and $m<n$, is there a closed form for this division? $\frac{\binom{n-m-1}{k}}{ \binom{n-1}{k}}$.
I ended up with $\frac{(n-m-1)!}{(-nk+mk+k)!}\times \frac{(-nk+k)!}{(n-1)!}$ and couldn't simplify it more. The case where $m=1$ I remember I found it $1-\frac{k}{n-1}$.
Your arithmetic after cancelling the factors of $k!$ seems to have gone astray.
$$\begin{align*} \frac{\binom{n-m-1}k}{\binom{n-1}k}&=\frac{(n-m-1)!k!(n-1-k)!}{k!(n-m-1-k)!(n-1)!}\\ &=\frac{(n-m-1)!(n-1-k)!}{(n-m-1-k)!(n-1)!}\\ &=\frac{(n-m-1)^{\underline k}}{(n-1)^{\underline k}}\\ &=\prod_{i=1}^k\frac{n-m-i}{n-i}\\ &=\prod_{i=1}^k\left(1-\frac{m}{n-i}\right) \end{align*}$$
Here $x^{\underline k}=x(x-1)(x-2)\ldots(x-k+1)$ is a falling factorial.