Prove by induction that $1 \cdot 1!+2 \cdot 2!+\ldots+n \cdot n!=(n+1)!-1$

Question

I am stuck at the inductive step. After much effort I arrived at $2(k+1)!+k(k+1)!-1$ but don't know how to go from there to $((k+1)+1!)-1$. I realise that I must express it in terms of $k+1$, but don't know how to manipulate the expression any further. Any help would be appreciated.

Answer 1

You have a commont factor $ (k+1)! $, $$ 2(k+1)! + k(k+1)! - 1 = (k+2)(k+1)! - 1 = (k+2)! -1. $$

Answer 2

Say we have $1.1!+2.2!…+(n-1).(n-1)!=n!-1$ for $n$. Then we will check it out for $n+1$. Since $(n+1)!= (n+1).n!=n.n!+n!$ then we have the equality $ (n+1)!-1=n.n!+n!-1$ and with induction hypothesis we get the last step $(n+1)!-1=n.n!+(n-1).(n-1)!+\ldots+1.1!$