Question asks to simplify:
$$\frac{^n\mathrm{C}_k}{^n\mathrm{C}_{k-1}}$$
I have a few steps but not sure if its correct.
$$\begin{align*}\frac{(n)!}{(n-k)!(k)!} \bigg/ \frac{(n)!}{(n-k-1)!(k-1)!} & = \frac{(n-k+1)!(k-1)!}{(n-k)!k!} \\ & = \frac{(n-k+1)!(k-1)k!}{(n-k)!k!}\end{align*}$$
Do the $k!$'s cancel out? Because I cancelled them
then I finally get $\dfrac{(n-k+1)!(k-1)}{(n-k)!}$
$$\frac{^n\mathrm{C}_k}{^n\mathrm{C}_{k-1}} = \frac{n!}{(n-k)!k!} \cdot \frac{(n-(k-1))!(k-1)!}{n!} = \frac{(n-k+1)\color{blue}{(n-k)!}\color{red}{(k-1)!}}{\color{blue}{(n-k)!} k\color{red}{(k-1)!}} = \frac{n-k+1}{k}$$