Number of all possible different combinations of a non-standard card deck.

Father and son play a game with a 52-card deck of cards. Each card of this deck has a number from 1 to 13 on it and there are 4 cards exactly the same for each number.

In how many different ways can these cards be placed in the deck?

My effort: I first began by considering the simpler problem where there are only 4 exact same aces, other than that the deck would be standard (2 colors, 4 symbols, 52 cards) except for the aces. To find all the possible different ways this specific deck can be arranged, I thought of considering the 4 aces as distinct ones, and then dividing by 4! since that's the number of different ways 4 items can be placed without repetitions.

Having solved the relaxed problem, I thought that by induction that the answer to the original question would be $$\frac{52!}{(4!)^{13}}$$.

Is my answer correct? Am I missing something big? Any help would be greatly appreciated.

To begin with, you can place the aces in $$52 \choose 4$$ ways. Then for each of those, you can place the twos in $$48 \choose 4$$ ways. And for each of those you have $$44 \choose 4$$ ways to place the threes. Expanding that out, we see that the aces, twos, and threes can end up arranged within the final deck in $$\begin{eqnarray} {52\choose 4}{48\choose 4}{44\choose 4} &=& \frac{52!}{(52-4)! 4!} \frac{48!}{(48-4)! 4!} \frac{44!}{(44-4)! 4!} \\ &=& \frac{52!}{(44-4)! (4!)^3} \end{eqnarray}$$ ways.
Proceeding in this way all the way up to the kings, for which there will always remain only $${4\choose 4}=1$$ way to place them, we also note that the product will continue to telescope—each subsequent factor’s numerator cancels with its predecessor’s denominator. So the answer is, as you suggest $$52!/(4!)^{13}$$.