Prove by induction: $1(1!)+\cdots + n\cdot n!$ = (n+1)! - 1

Question

Induction step. $1(1!) + ... + n(n!) = (n+1)! - 1$
$1(1!) + ... + n(n!) + (n+1)(n+1)! = (n+1)! - 1 + (n+1)(n+1)!$
So, I don't understand how to get $(n+2)! - 1$ from $(n+1)! - 1 + (n+1)(n+1)!$.
Please, explain it step by step.

Answer 1

Do some factoring: \begin{align*} (n+1)!−1+(n+1)(n+1)! &= [(n+1)(n+1)! + 1(n+1)!] - 1 \\ &= [(n+1) + 1](n + 1)! - 1 \\ &= (n + 2)(n + 1)! - 1 \\ &= (n + 2)! - 1 \\ \end{align*}

Answer 2

For $n=1$ the equality is trivial: $$ 1(1!)=1 =(1+1)!-1$$ Then, we have the basis of induction.

Now, supose that the equality is true for $n=k$, i.e., $$1(1!)+...k(k!)=(k+1)!-1 . (*)$$

We need to show that is true for $n=k+1$, i.e., $$1(1!)+...+(k+1)(k+1)!=(k+2)!-1. $$ Indeed, summing $(k+1)(k+1)!$ in (*) we obtain $$1(1!)+...+k(k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=(k+1)![1+(k+1)]-1=$$ $$=(k+1)![k+2]-1 =(k+2)!-1 .$$

So, the result follows by the principle of mathematical induction.