How does $(n+1)! - 1 + (n+1)(n+1)! = (n+1)!(1+n+1) - 1$?

Question

How does $(n+1)! - 1 + (n+1)(n+1)! = (n+1)!(1+n+1) - 1$?

I cannot figure this out, help. This deals with Mathmatical induction problem. I've tried factoring but it doesn't work out.

Answer 1

$(n+1)!-1+(n+1)(n+1)! = (1\cdot(n+1)!+(n+1)(n+1)!)-1 = (n+2)(n+1)!-1 = (n+1)!(1+n+1)-1.$

Answer 2

Notice that showing $$(n+1)! - 1 + (n+1)(n+1)! = (n+1)!(1+n+1)-1$$ is equivalent to showing $$(n+1)! + (n+1)(n+1)! = (n+1)!(1+n+1)\text{.}$$ (Simply add $1$ to both sides.) Now $$(n+1)!+(n+1)(n+1)! = (n+1)!(1+(n+1)) = (n+1)!(1+n+1)$$ after factoring, so we have solved the problem. If you are picky about how this looks, subtract $1$ to both sides.

Answer 3

Set $K=(n+1)!$; then your expression becomes \begin{align} (n+1)! - 1 + (n+1)(n+1)! &=K-1+(n+1)K\\[6px] &=K(1+n+1)-1\\[6px] &=(n+2)K-1\\[6px] &=(n+2)\cdot(n+1)!-1\\[6px] &=(n+2)!-1 \end{align} Note that the second line is exactly what you wanted to prove; I went on to show a further simplification.