An equality with factorials

Question

Prove that $$2\times 2! + 3\times 3!+4\times 4! +....+n\times n!=(n+1)!-2$$ I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction. I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$ But I did not get any thing useful.

Answer 1

Hint: $$\sum_{k=2}^n k\cdot k!=\sum_{k=2}^n (k+1-1)\cdot k!=\sum_{k=2}^n (k+1)!-\sum_{k=2}^n k!$$

Answer 2

$2\times2!=3!-2!$

$3\times3!=4!-3!$

...

$n\times n!=(n+1)!-n!$

Add them all up then $3!,4!,...n!$ cancel out.

Therefore LHS$=(n+1)!-2$