How to get $(n+1)!(n+2) = (n+2)!$ I mean it makes sense when I look at it that the two are equal, but I don't entirely understand how you get from one to the other - I presume there's some basic algebra involved - but I'm not sure what it is, how to figure it out, could someone explain this concept to me? I understand it somewhat relates to telescopic series, but I feel like if I don't understand this basic concept, I won't really get the rest of it.
$$(n+1)\cdot(n+1)!=(n+2-1)\cdot(n+1)!=(n+2)\cdot(n+1)!-(n+1)!=?$$
(n+2)!=(n+1)!(n+2)

(n+1)!((n+1)+1)=(n+2)!

(n+1)!(n+1)+(n+1)!=(n+2)!

(n+1)!(n+1)=(n+2)!-(n+1)!

hmmm..... where was i wrong on this?!
Edits:
(n+1)!(n+1)=(n+1)!(n+2)-(n+1)!

(n+1)!(n+1)=(n+1)!(n+1)

hmmm.... yah. 

also: the fixed title: lets say n=3, then (3+1)!(3+2)=(3+2)!

(3+1)!=1*2*3*4

(3+2)=5

(3+2)!=1*2*3*4*5
get it?