Question 1081665
By definition, 


n! = n*(n-1)*(n-2)*...*3*2*1


Whatever value you pick for n, you count your way down til you get to 1. All of the integers in the countdown are multiplied.


----------------------------------------------


Using similar logic, the expression (n+5)! can be expanded as such


(n+5)! = (n+5)*(n+4)*<font color=red>(n+3)*(n+2)*(n+1)*...*3*2*1</font>


and that can be turned into


(n+5)! = (n+5)*(n+4)*<font color=red>(n+3)!</font>


notice how the red portions match up. They are equal expressions. In other words,


(n+3)! = (n+3)*(n+2)*(n+1)*...*3*2*1


----------------------------------------------


Let's use that idea to simplify


{{{((n+5)!)/((n+3)!)=((n+5)(n+4)(n+3)!)/((n+3)!)}}}


{{{((n+5)!)/((n+3)!)=((n+5)(n+4)*highlight((n+3)!))/(highlight((n+3)!))}}}


{{{((n+5)!)/((n+3)!)=((n+5)(n+4)*cross((n+3)!))/(cross((n+3)!))}}}


{{{((n+5)!)/((n+3)!)=(n+5)(n+4)}}}


{{{((n+5)!)/((n+3)!)=n^2+9n+20}}}


----------------------------------------------


Final Answer: {{{n^2+9n+20}}}