TITLE: Proving $\sum_{k=1}^n k\cdot k! = (n+1)!-1$ without using mathematical Induction.
QUESTION [11 upvotes]: Possible Duplicate:
Summation of a factorial 

This equation is given:
$$
1\cdot1! + 2\cdot2! + 3\cdot3! + \ldots  + n\cdot n! = (n+1)! - 1
$$
I've solved it using mathematical induction but I'm curious what could be the other possible ways to prove it.

REPLY [10 votes]: Both sides count the number of nontrivial permutations of $n+1$ objects.  By "nontrivial" I mean "not the identity".
RHS: The total number of permutations is $(n+1)!$, of which 1 is trivial.
LHS: Let the objects be $0,1,\dotsc,n$.  Every nontrivial permutation moves at least one number.  Let $k$ be the greatest number which is moved.  Since the greater numbers are not moved, the permutation must move $k$ into one of the $k$ spots to its left.  The $k$ numbers $0,1,\dotsc,k-1$ can then be arranged in the remaining spots in $k!$ ways.  So that's $k\cdot k!$ permutations in which $k$ is the greatest number moved; sum over $k$ to get the LHS.