Answer
$a_n=\frac{(n+1)!}{n+3}$
Work Step by Step
$\frac{2!}{4},\frac{3!}{5},\frac{4!}{6},\frac{5!}{7},\frac{6!}{8},...$
$\frac{(1+1)!}{1+3},\frac{(2+1)!}{2+3},\frac{(3+1)!}{3+3},\frac{(4+1)!}{4+3},\frac{(5+1)!}{5+3},...$
$a_n=\frac{(n+1)!}{n+3}$
