Prove this factorial fractional equivalence

Question

I'm supposed to prove the following: $$ \sum_{i=0}^{n-1}\dfrac{(2n-2-i)!}{(n-1-i)!} = \sum_{i=0}^{n-1} \dfrac{(n+i-1)!}{i!}$$

I would greatly appreciate it if people could please take the time to clarify how to do this.

is there any way I can rewrite the first term to get to the second one? I am familiar with the properties but I couldn't make any use of them so far..

EDIT: made a mistake

Answer 1

Hint: Consider $$ \frac{\left(2n-2-i\right)!}{\left(n-1-i\right)!} $$ and set $i=n-1-j$ to get $$ \frac{\left(2n-2-\left(n-1-j\right)\right)!}{\left(n-1-\left(n-1-j\right)\right)!}=\frac{\left(n+j-1\right)!}{j!}. $$