Card dealing question

Q: In how many ways can a pack of 52 cards be dealt to 13 players, 4 to each, so that one player has a card from each suit but no one else has cards from more than one suit?

A: Three steps:
1. pick one player, pick one card from each suit
==>13^5
2. put 12 remaining players into 4 different suits to make three players per suit
==> 12!/(9!*3!) * 9!/(6!*3!) * 6!/(3!*3!) = 12!/(3!)^4
3. put 12 remaining cards in each suit to 3 players, separately
==> (12!/*8!*4!) * 8!/(4!*4!))^4 = (12!)^4 / (4!)^12

so final answer is: (13!)^5 / (4!)^12 / (3!)^4