Sum the series $(1^2+1)1! + (2^2+1)2! + (3^2+1)3! \cdots + (n^2+1)n!$ Problem:
Sum the series: $$(1^2+1)1! + (2^2+1)2! + (3^2+1)3! \cdots + (n^2+1)n!$$
Source: A book on algebra.I came across this interesting looking series and decided to tackle it. 
My try :
All I have tried is taking the $r^{th}$ term and summing it, but in vain:
$$ T_r = (r^2+1)r!$$
$$T_r = r^2\cdot r! + r!$$
Now I don't know how to sum either of these terms.I'm not familiar with uni level math as I'm a high school student. All help appreciated!

$$
S=\sum_{r=1}^n\left( (r+2)!-3(r+1)!+2r!\right)\\
=(n+2)!+(n+1)!-3(n+1)!-3\cdot2+2\cdot1+2\cdot2\\
=n(n+1)!
$$
