Combinatorial proof involving sum of factorials I need help with this combinatorial proof:
$1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1$
So far I came up with this:
Let S be a set of numbers $1, 2, \ldots, n+1$
So LHS could be: How many permutations of length $k$ there is over set $S$ such that number $1$ is not the first in permutation. $k$ can be $2, 3,\ldots, n+1$. But the RHS gives me trouble...
Thanks.
(Sorry for the poor formating, I am not good with HTML)
$n\cdot n!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ without $n+1$ being first. 

$(n-1)\cdot (n-1)!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ with $n+1$ being first and $n$ not second. 

$(n-2)\cdot (n-2)!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ with $n+1$ being first and $n$ second and $n-1$ not third. 

Etc...

The RHS is all the ways to order $\{1,2,3,\ldots,n+1\}$ without all the elements in descending order.
hint: $n\cdot n! = (n+1)!-n!$. Can you continue?