Find the sum of the series:
. . $\displaystyle 1(1!) + 2(2!) + 3(3!) + \cdots + n(n!)$
I don't know where it was solved, but even without (explicit) induction:
$\displaystyle
k\left( {k!} \right) = \left( {k + 1} \right)k! - k! = \left( {k + 1} \right)! - k!
$
So summing for k = 1 till n gives (n+1)!-1 since all intermediate terms cancel out.