What is the value of sum of k!*k, for k = 1 to n?

Required to find the sum k!*k for k = 1 to k =
n.

Let Sn be the sum  of the n terms of the series.. We
write the actual series for k = t to k= n:

Sn =
1!/*1+2!*2+3!*3+4!*4 +....+(n-1)!.(n-1)+n!*n-----(1).

We
take the k th term  k!*k= (1*2*3*....r)*k , k being an integer  of natural
number.

So r!*r = (1*2*3*4*...*k)(k+1 -1) = (k+1)! -
k!....(2).

So now we rewrite each term of on the right side
of (1) as in (2):

Sn =
(2!-1!)+(3!-2!)+(4!-3!)+(5!-4!)+.......[(n-1)!-(n-2)!]+
[(n!-(n-1)!+[(n+1)!-n!].-----(3).

The right side of (3)
could be rearranged as below:

Sn =
-1!+{(-2!+2!)+(-3!+3!)+(-4!+4!)+......(-n!+n!)}+(n+1)!.

Sn
= -1!+(n+1)!.

Sn = (n+1)!-1, as 1!=
1.

Therefore the sum of k!*k for k= 1 to k =
n is (n+1)!-1.