Permutations and Combinations based probblem Find the value of the expression:
$$
1+1\times1!+2\times2!+3\times3!+.....+n\times n!
$$
It is a problem based on the concept of permutations and combinations I don't have a perfect idea to solve this. As I use a very long method which is worthless. Suggest me some easier as well as shorter method. Thanks in advance 
 A: Here’s a combinatorial argument. We’ll count the permutations of $[n+1]=\{1,\ldots,n+1\}$ that are not the identity permutation $123\ldots n(n+1)$ in two ways. There are clearly $(n+1)!-1$ such permutations.
For each such permutation $\pi=p_1p_2\ldots p_{n+1}$ let $m(\pi)=\max\{k:p_{k+1}\ne k+1\}$; $m(\pi)$ exists, since $\pi$ is not the identity permutation. Note that $m(\pi)\ge 1$: if $p_1\ne 1$, then necessarily there is some $k>1$ such that $p_k\ne k$. For $k=1,\ldots,n$ let $P_k$ be the set of non-identity permutations $\pi$ of $[n+1]$ such that $m(\pi)=k$; clearly
$$(n+1)!-1=\sum_{k=1}^n|P_k|\;.$$
For all $\pi\in P_k$ we have $p_{k+1}\ne k+1$ and $p_i=i$ for $k+1<i\le n+1$, so $p_1p_2\ldots p_{k+1}$ is a permutation of $[k+1]$. The only restriction is that $p_{k+1}\ne k+1$, so it can be any permutation of $[k+1]$ that does not have $k+1$ as a fixed point. There are therefore $k$ possible values for $p_{k+1}$, and the remaining $k$ members of $[k+1]$ can be permuted arbitrarily, so there are altogether $k\cdot k!$ such permutations, and $|P_k|=k\cdot k!$. Thus,
$$\sum_{k=1}^nk\cdot k!=\sum_{k=1}^n|P_k|=(n+1)!-1\;.$$
A: We have
$$
\sum_{i = 0}^n i\times i! + \sum_{i = 0}^n i! = \sum_{i=1}^n(i+1)!
$$
so
$$
\sum_{i = 0}^n i\times i! = (n+1)! - 1,
$$
so your expression is precisely $(n+1)!$.
