Question 29757
show that 1(1!)+2(2!)+3(3!)+...+n(n!)= (n+1)!-1 for all positive integers n.
LET SN =  1(1!)+2(2!)+3(3!)+...+n(n!)
CONSIDER THE EQN....
(N+1)!-N! = N!(N+1-1)= N*N!...PUT N =1,2,3 ETC....AND ADD
N=1.........2!-1! = 1*1!
N=2.........3!-2! = 2*2!
N=3.........4!-3! = 3*3!
........................
N=N-1....N!-(N-1)! = (N-1)*(N-1)!
N=N......(N+1)! - N!= N*N!
----------------------------------ADDING ALL 
...........(N+1)!-1!=SN