Combinatorics problem involving series summation Can you give a combinatorial proof of $1\cdot1! + 2\cdot2! ....+ (n-1)(n-1)! = n!-1$
All permutations of $\{1,2,\ldots,n\}$ objects (except the identity) can be obtained first swapping $k$ with its desired value and then permuting the lower $k-1$ objects, where $k$ (with $2\le k\le n$) is uniquely determined as the highest non-fixed position.
$$\sum_{i=1}^{n-1}i\cdot i !=\sum_{i=1}^{n-1}(i+1-1)\cdot i !=\sum_{i=1}^{n-1}(i+1)\cdot i! -\sum_{i=1}^{n-1}i !=\sum_{i=1}^{n-1}(i+1)! -\sum_{i=1}^{n-1}i !=\sum_{i=2}^{n}i !-\sum_{i=1}^{n-1}i !=n!-1!=n!$$