I have a question is it 1*1!+2*2!+...n*n! or 1.1!+2.2!+...n.n! where "." is a decimal because if its the second case you cannot solve the equation its a math error and the answer would be E.

But I'm assuming its the first option

You can pick some values for n to solve this question.
Let n=3

We need to find the value of 1*1!+2*2!+3*3!=23

Now substitute n=3 in the answer choices and see which choice gives you the answer as 23

The only option that works is (n+1)!-1

So I guess the answer should be C.

ans = 1.1! + 2.2! + 3.3! . . . . n.n!
add 1 both sides

ans + 1 =
1 + 1.1! + 2.2! + 3.3! . . . . n.n!
2! + 2.2! + 3.3! . . . . n.n!
2! (1+2) + 3.3! . . . . n.n!
3! + 3.3! . . . . n.n!
3! (1+3) . . . . n.n!
4! ... n.n!
at the end you will get (n+1)!

so ans = (n+1)! - 1