Trying to understand an exercise using factorials with induction Exercise:
Prove that (n + 1)! - n! = n(n!) for any n $\ge$ 1
Given Answer:
I will skip the basic step since I understand that part.
(n + 2)! - (n + 1)! = (n + 1)!(n + 2) - n!(n + 1) I understand this line 
But, I don't understand starting at this next line

= n(n + 1)! + 2(n + 1)! - n(n!) - n!
= n(n + 1)! + (n + 1)! - n(n!) + (n + 1)! - n! 
= n(n + 1)! + (n + 1)! - n(n!) + n(n!) 
= n(n + 1)! + (n + 1)! 
= (n + 1)!(n + 1) 
If anyone can help me to understand this then that would be great!
Thanks, 
Tony
 A: $(n+2)! = (n+2)(n+1)! = n(n+1)! + 2(n+1)!$, by the distributive property.

The inductive step would be to assume that $(n + 1)! - n! = n(n!) $ holds for some $n > 1$.
We then want to show that $\color{blue}{(n+2)!} -\color{red}{(n+1)!} = (n+1)(n+1)!$
We begin by rearranging...
$$\color{blue}{(n+2)!} -\color{red}{(n+1)!} =\color{blue}{(n+2)}(n+1)! - \color{red}{(n+1)}(n!)$$
$$=\color{blue}{n}(n+1)! + \color{green}{2}(n+1)! - \color{red}{n}(n!) -\color{red}{1}(n!) $$
Notice that I changed the blue $2$ into green. That is so we can have $$=\color{black}{n}(n+1)! + \color{green}{1}(n+1)! - \color{black}{n}(n!) + \color{green}{1}(n+1)! -\color{black}{1}(n!) $$
The two terms on the right are $(n+1)! - n!$, which equals $\color{red}{n(n!)}$ by hypothesis. So now we have...
$$=\color{black}{n}(n+1)! + \color{black}{1}(n+1)! - \color{black}{n}(n!) + \color{red}{n(n!)} $$
The two terms on the right cancel. $$=\color{blue}{n}(n+1)! + \color{blue}{1}(n+1)!$$
Using the distributive property (factoring out a $(n+1)!$, if you will) ...
$$\color{blue}{(n+1)}(n+1)!$$
So the claim holds for $n+1$.
