Alright, so, two things,
k! = (k)(k-1)(k-2)(k-3)(k-4)(k-5)...etc
(k+3)! = (k+3)(k+3-1)(k+3-2)(k+3-3)(k+3-4)(k+3-5)
now you'll notice that, simplifying, (k+3-3) = k, and (k+3-4) = (k-1), and (k+3-5) = (k-2), etc.
so in effect, (k+3)! has all of the terms as k!, with the except of the first three terms of (k+3)!, so you can divide them out
hopefully that explains it well enough. if not lemme know