Sum $\sum\limits_{k=1}^n (k^2+1)\cdot k!$ I could not compute the sum of the following series-

$$\displaystyle\sum_{k=1}^n (k^2+1)\cdot k!$$
Please tell me how to proceed in the above question.
Any help is appreciated.

It can be proven by induction that
$$\sum_{k=1}^n (k^2+1)k! = n(n+1)!$$
This clearly holds for $n=1$. Not suppose it holds for $n$. Then
$$\sum_{k=1}^{n+1}(k^2+1)k! = ((n+1)^2+1)(n+1)!+\sum_{k=1}^{n}(k^2+1)k!$$
$$=((n+1)^2+1)(n+1)! + n(n+1)!$$
$$= (n^2+3n+2)(n+1)! = (n+1)(n+2)(n+1)! = (n+1)(n+2)!$$
Thus completing the proof. 
