Partial sum using telescopic series The summation of $(n^2+1)\cdot n!$ using to $N$ using telescopic series.
Question: Find a closed expression in $n$ for:
\begin{gather}
\text{Let } u_n = (n^2+1)\cdot n! \\
\sum_{n=1}^{N}u_n = \sum_{n=1}^{N}(n^2+1)\cdot n!\\
\end{gather}
My Attempt
I tried to split $(n^2+1)\cdot n!$ into two different terms that would telescope. 
First try
\begin{gather}
(n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\
= (n+1)\cdot (n+1)! - 2n\cdot n!\\
\sum_{n=1}^{N}(n+1)\cdot (n+1)! - 2n\cdot n!\\
(2)\cdot(2)! - 2(1)\cdot(1)!\\
(3)\cdot(3)! - 2(2)\cdot(2)! \\
\vdots \\
(n+1)\cdot (n+1)! - 2(n)\cdot (n)!\\
\end{gather}
The $2$ on the second term is quite annoying and needless to say it does not work. 
Second try
\begin{gather}
(n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\
= (n+1)\cdot (n+1)! - 2n\cdot n!  =  (n+1)\cdot(n+1)! -2[(n+1)! - (n)!]\\
\end{gather}
Didn't work.
Try #3
\begin{gather}
(n+1)\cdot(n+1)! -2[(n+1)! - (n)!] = (n+1)!\cdot(n-1)+2n!\\
\end{gather}
This one didn't help at all.
Could anyone please suggest a hint to arrive at the right telescoping expression?

The key is that $$(n^2+1)\cdot n! = n\cdot(n+1)! - (n-1)\cdot n!.$$
Thus $$\sum_{n=1}^Nu_n=(N\cdot(N+1)!-(N-1)\cdot N!) + ((N-1)\cdot N! - (N-2)\cdot (N-1)!) +\\+ \cdots + (1\cdot2!-0\cdot1!) = N\cdot(N+1)!.$$
