How to find the Summation of series of Factorials? $$1\cdot1!+2\cdot2!+\cdots+x\cdot x! = (x+1)!−1$$
I don't understand what's happening here. The given sum of factorials is generalized into a single term. Could somebody please help me finding the logic behind this series.
Thanks in advance!!
 A: $$
n!\cdot n \quad = \quad n!\Big((n+1) - 1\Big) \quad = \quad  \big(n!(n+1) \big) - \big( n!\cdot1 \big) \quad = \quad (n+1)! - n!.
$$
Therefore
\begin{align}
& 1!1+2!2 + 3!3 + 4!4 + 5!5 \\[10pt]
= {} & \Big(2!-1!\Big) + \big( 3!-2!\Big) + \Big(4!-3!\Big) + \Big(5!-4!\Big) + \big(6!-5!\Big)
\end{align}
Now notice that all terms cancel except the biggest one and the smallest one.
A: Michael Hardy’s computational proof is the simplest, but there is also a reasonably straightforward combinatorial argument.
There are $(n+1)!-1$ permutations of the set $[n+1]=\{1,2,\ldots,n+1\}$ other than the increasing permutation $\langle 1,2,\ldots,n+1\rangle$. Let $P$ be the set of all such permutations; we’ll now calculate $|P|$ in another way.
For $k=1,\ldots,n$ let $P_k$ be the set of permutations $\langle p_1,p_2,\ldots,p_{n+1}\rangle\in P$  such that $p_k\ne k$, and $p_\ell=\ell$ for $1\le\ell<k$. If $\langle p_1,\ldots,p_{n+1}\rangle\in P_k$, $p_k$ cannot be any of the positive integers less than or equal to $k$, so there are $(n+1)-k$ possible choices for $p_k$. Once $p_k$ has been set, $\langle p_{k+1},\ldots,p_{n+1}\rangle$ can be any permutation of the remaining $(n+1)-k$ members of $[n+1]$, so 
$$|P_k|=(n+1-k)(n+1-k)!\;.$$
Thus, 
$$(n+1)!-1=\sum_{k=1}^n|P_k|=\sum_{k=1}^n(n+1-k)(n+1-k)!=\sum_{\ell=1}^n\ell\cdot\ell!\;,$$
where in the last step I just set $\ell=n+1-k$.
