Finding summation $1+\sum\limits_{r=0}^{18} (r(r+2)+1)r! = k!$

Question

It is being given that :

$$1+\Sigma_{r=0}^{18} (r(r+2)+1)r! = k!$$

We need to find $k$.

What I found was

$$1+\Sigma_{r=0}^{18} (r(r+2)+1)r!$$

$$=1+\Sigma_{r=0}^{18} (r+1)^2r!$$

$$=1+\Sigma_{r=0}^{18} (r+1)\cdot(r+1)!$$

Now how do I proceed?

Answer 1

Hint:

$$(r+2-1)(r+1)!=(r+1)(r+1)!$$

$$(r+2)!-(r+1)!=(r+1)(r+1)!$$