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

(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

• Factor out the $(n+1)!$ from the two terms that have it. So, use commutativity of addition and subtraction to move the $-1$ to the end, then you have $1\times (n+1)!+(n+1)\times (n+1)!-1$. – InterstellarProbe May 2 at 20:06
• Take out common factor $(n+1)!$ – lab bhattacharjee May 2 at 20:07

$$a\times c + b\times c = (a+b)\times c$$
$$c = 1\times c$$
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:
$$(n+1)! + (n+1)\times (n+1)! = 1\times (n+1)! + (n+1)\times (n+1)! = (1+n+1)\times (n+1)!$$