I am having trouble understanding some of the algebraic concepts used here. In fact, the entire thing to me makes sense, except for the second red line. I don't understand how the diagonal swap happened there. Why did the (n-1)! on the LHS swap with the (n-3)! on the RHS?
After that all makes sense to me, I just don't get this one bit.
Thanks,
They multiply the both sides by $\frac{(n-3)!}{(n-1)!}$.
$$\frac{4(n!)}{(n-3)!}=\frac{5(n-1)!}{(n-1-3)!}$$$$\Rightarrow \frac{4(n!)}{\color{red}{(n-3)!}}\times \frac{\color{red}{(n-3)!}}{(n-1)!}=\frac{5\color{green}{(n-1)!}}{(n-1-3)!}\times\frac{(n-3)!}{\color{green}{(n-1)!}}$$ $$\Rightarrow \frac{4(n!)}{(n-1)!}=\frac{5(n-3)!}{(n-4)!}$$