How to prove $\sum_{i=1}^n i.i! = (n+1)!-1$ with mathematical induction?

Question

I'm trying to prove $$ \sum_{i=1}^n i.i! = (n+1)!-1 $$ with mathematical induction. The first step I did after prove it for 1 was: $$ \sum_{i=1}^{n+1} i.i! = (n+2)!-1= (n+2).(n+1)!-1 $$ but I can't do anything more.

Answer 1

The equality is trivial for $n=1.$ Assume it is true for $n,$ then $$\begin{align*}\sum_{i=1}^{n+1} i\cdot i!&=\sum_{i=1}^n i.i!+(n+1).(n+1)!\\&=(n+1)!-1+(n+1)(n+1)! \qquad \text{(induction hypothesis)}\\ &=(n+2)(n+1)!-1\\&=(n+2)!-1.\end{align*}$$