Induction proof for n > 0

Question

Prove using induction the following: for n > 0,

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

I'm not very good at proving proofs with the induction method, help would be greatly appreciated

Answer 1

For the inductive step, assume that $1 \cdot 1! + \cdots + n \cdot n! = (n+1)!-1$. Then, \begin{align*} 1 \cdot 1! + \cdots + n \cdot n! + (n+1)\cdot (n+1)! &= (n+1)\cdot(n+1)! + (n+1)! - 1 \\ &= (n+1+1)\cdot (n+1)!-1. \end{align*} I leave the rest of the step and the base case to the reader.

Answer 2

Can you show that $$(n+1)!-1+(n+1)\cdot(n+1)!=(n+2)!-1$$ (and show why that matters)? If so, you've basically finished the proof, aside from a few details, like the base step.

Answer 3

(n+2)! - 1 = (n+2)(n+1)! - 1 = (n+1+1)(n+1)! - 1 = (n+1)(n+1)! + [(n+1)! - 1] = (n+1)(n+1)! + [(n)(n!) + ... + (2)(2!) + (1)(1!)]