Proving $1\times 1! + 2\times 2! + 3\times 3! +\cdots.+n\times n! = (n+1)! -1 $

Question

Prove that: $$1\times 1! + 2\times 2! + 3\times 3! +\cdots+n\times n! = (n+1)! -1 $$

My attempt :-

Let

S= $1\times 1! + 2\times 2! + 3\times 3! + ...+n\times n! = (n+1)! -1 $ = $1\times 1! + 4\times 1! + 9\times 2! + 16\times 3! + ...+ n^2 \times (n-1)!$

Now i wanna relate this sum with the original sum to get it but i do not know how can i do it ?

Answer 1

Hint: Note that

$$\begin{equation}\begin{aligned} n(n!) & = (n + 1 - 1)n! \\ & = (n+1)n! - n! \\ & = (n+1)! - n! \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This shows the sum forms a telescoping series.