Factorials

(pq)! = (pq)(pq-1)...(1)
note the above expansion will also contain q! ~~~~~~~(X)
also the above expansion has pq terms
so the terms can be grouped into q groups with p CONSECUTIVE terms each
now you must be knowing product of k consecutive terms is a multiple of k!
so p! divides each group.
and there are q groups in all so (p!^q) divides (pq)! ~~~~~~~~~~(Y)
from (X) and (y) we get
(p!^q)q! divides (pq)!
the second part can be proved similarly

this proof is on the lines of prophet sir's proof here
http://targetiit.com/iit-jee-forum/posts/permutation-combination-10320.html

This proof holds only when p and q are relatively prime whereas a combinatorial argument will easily show that this is true for any pair (p,q)