Probability of deck of cards such that each person receives one ace

There are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$ ways to distribute $13$ cards to each of four people.

There are $4!$ ways to distribute the aces so that each person receives one and $$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$ ways to distribute the remaining cards so that each person receives twelve of them. Hence, the desired probability is \begin{align*} \frac{4!\dbinom{48}{12}\dbinom{36}{12}\dbinom{24}{12}\dbinom{12}{12}}{\dbinom{52}{13}\dbinom{39}{13}\dbinom{26}{13}\dbinom{13}{13}} & = \frac{4! \cdot \dfrac{48!}{12!36!} \cdot \dfrac{36!}{12!24!} \cdot \dfrac{24!}{12!12!} \cdot \dfrac{12!}{12!0!}}{\dfrac{52!}{13!39!} \cdot \dfrac{39!}{13!26!} \cdot \dfrac{26!}{13!13!} \cdot \dfrac{13!}{13!0!}}\\[2mm] & = \frac{4! \cdot \dfrac{48!}{12!12!12!12!}}{\dfrac{52!}{13!13!13!13!}}\\[2mm] & = \frac{4!48!}{12!12!12!12!} \cdot \frac{13!13!13!13!}{52!}\\[2mm] & = \frac{4!48!13^4}{52!}\\[2mm] & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}

Let's compare this solution with the approach of your author. As you stated, there are $\binom{52}{4}$ ways to choose the four positions occupied by the aces in the deck. Since each person receives $13$ cards, there are $13$ possible places for the position of the ace in each person's hand. Hence, the desired probability is $$\frac{13^4}{\dbinom{52}{4}}$$

All we really care about is the placement for the aces; not where the other cards may be in the deck, nor even the suits of the aces.

So let us take 52 blank cards and 4 stickers with 'ace' written on them.   To 'deal the cards', place the blank cards in for lines of 13, then unbiasedly select four from them and put a sticker on each.

How may ways are there to stick the aces on 4 different cards in the deck of 52? That is a selection of 4 from 52. $$\binom {52}4$$

How many ways are there to do this so that each ace is stuck on one card in each from the four lines of 13? That is a selection of 1 from the first line, 1 from the next line, and so forth. $$\binom{13}1\binom{13}1\binom{13}1\binom{13}1$$

Divide and calculate.$$\dfrac{13^4}{\dbinom{52}{4}}$$