TITLE: An equality with factorials
QUESTION [8 upvotes]: 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.

REPLY [15 votes]: $2\times2!=3!-2!$
$3\times3!=4!-3!$
...
$n\times n!=(n+1)!-n!$
Add them all up then $3!,4!,...n!$ cancel out.
Therefore LHS$=(n+1)!-2$