Can someone explain how does step 1 render to step 2?

Question

(1)$(n+1)!−1+(n+1)×(n+1)!$

(2)$=(1+n+1)×(n+1)!−1$

(3)$=(n+2)×(n+1)!−1 $

(4)$=(n+2)!−1$

I understand how step 4 derived from 3, but I am confused on how does step 2 derived from step 1? thank you

Answer 1

$a\times c + b\times c = (a+b)\times c$

$c = 1\times c$

Now, use the above with $1$ in place of $a$, with $(n+1)$ in place of $b$ and $(n+1)!$ in place of $c$ to get:

$(n+1)! + (n+1)\times (n+1)! = 1\times (n+1)! + (n+1)\times (n+1)! = (1+n+1)\times (n+1)!$