TITLE: Combinatorial Proof: $n! = 1+\sum\limits_{r=1}^{n-1}( r\cdot r!)$
QUESTION [7 upvotes]: To show it is true $r\times r!$ can be algebraically decomposed as $(r+1)!-r!$ 
But I am trying to think of a combinatorial proof. If $S$ is the set of all permutations of $\{1,2,\cdots ,n\}$ then I need a definition for the mutually non overlapping subsets $S_r$ whose union is $S$. So I will have $|S|=\sum|S_r|$. I am unable to think of such a decomposition.

REPLY [8 votes]: Hint: We are splitting up into the permutations which do not fix $1$, (There are ($(n-1)\times (n-1)!$) of these), and the ones that do (everything else).  To count the ones which fix $1$, we split into the $(n-2)\times(n-2)!$ permutations that do not fix $2$, and the left over number which do fix $2$.  Continuing in this way we get your sum.  (The one comes from the identity permutation)