An equality with factorials
Hint: $$\sum_{k=2}^n k\cdot k!=\sum_{k=2}^n (k+1-1)\cdot k!=\sum_{k=2}^n (k+1)!-\sum_{k=2}^n k!$$
$2\times2!=3!-2!$
$3\times3!=4!-3!$
...
$n\times n!=(n+1)!-n!$
Add them all up then $3!,4!,...n!$ cancel out.
Therefore LHS$=(n+1)!-2$