Factorial of any natural number $n$ can be expanded as a summation
$$n!=1+\sum_{i=1}^{n-1}(i\times i!)$$$$=1+1\times 1!+2\times 2!+3\times 3!+4\times 4!+..............+(n-1)\times (n-1)!$$ $\forall \ \ n\in N$
Although, I have derived & proved, by using induction, the above expansion of the factorial here on the basis of the generalized formula derived by me still I have doubt if there is another method to derive/prove.
Is there is an easier mathematical derivation/proof of the above expansion of factorial?
Any help will be greatly appreciated!
For fun, we give a combinatorial argument.
The numbers $1$ to $n$ are seated in a row, in that order. We count, in two different ways, the number of ways to reseat these numbers so that at least one number moves. The usual way of counting gives $n!-1$.
Maybe $1$ is the smallest number that moved. There are $n-1$ ways to choose its new position. For each of these ways, there are $(n-1)!$ ways to choose the positions of the remaining $n-1$ numbers, for a total of $(n-1)(n-1)!$.
Maybe $2$ is the smallest number that moved. There are then $n-2$ ways to choose where $2$ goes. For each of these ways there are $(n-2)!$ ways to seat the rest of the numbers, for a total of $(n-2)(n-2)!$.
Maybe $3$ is the smallest number that moved. There are then $n-3$ ways to choose where it goes. For each of these ways there are $(n-3)!$ ways to seat the rest of the numbers, for a total of $(n-3)(n-3)!$.
And so on. Finally, maybe $n-1$ was the smallest number that moved. It only had $1$ place to go, and the remaining number $n$ also has $1$, here called $1$, place to go.
That gives (backwards) a total of $\sum_{k=1}^{n-1}k\cdot k!$ permutations in which at least one number moves.