Manipulating factorials algebraically

Question

Hi I was wondering how to prove (if possible) that $\frac{(n+2)!-(n+1)!}{(n+1)!-n!}=\frac{(n+1)^2}{n}$.

I am just going into calculus 2 so I’m sure there is a way to prove this I’ve just not learned it yet.

Answer 1

$\frac{(n+2)!-(n+1)!}{(n+1)!-n!}=\frac{(n+2)(n+1)n!-(n+1)n!}{(n+1)n!-n!}=\frac{(n+2)(n+1)-(n+1)}{n+1-1}=\frac{(n+1)^2}{n!}$

Answer 2

$$\frac{(n+2)!-(n+1)!}{(n+1)!-n!}=\frac{(n+2)(n+1)n!-(n+1)n!}{(n+1)n!-n!}$$ Factor out $(n+1)n!$ from numerator and $n!$ from denominator. $$=\frac{(n+1)n!(n+2-1)}{n!(n+1-1)}=\frac{(n+1)^2}{n!}$$

So, $$\frac{(n+2)!-(n+1)!}{(n+1)!-n!}=\frac{(n+1)^2}{n!}$$