Your calculation treats $\heartsuit 3,\spadesuit 4,\clubsuit 5,\diamondsuit 6,\heartsuit 7$ and $\diamondsuit 6,\spadesuit 4,\heartsuit 7,\heartsuit 3\clubsuit 5$, for instance, as different hands, when in fact they’re the same set of five cards: the only difference between them is the order in which they were dealt. Since a given set of $5$ cards can be dealt in $5!$ different orders, you’re counting each hand of $5$ cards $5!$ times. To get rid of this overcounting, you must divide by $5!$. When you do, you get
$\frac{52!}{47!}\cdot\frac1{5!}=\frac{52!}{5!47!}=\binom{52}5\;,$
the number of ways of choosing an unordered set of $5$ cards from a $52$-card deck. Your $\dfrac{52!}{47!}$ is the number of ways to deal $5$ cards: it counts each of the $5!=120$ possible dealing orders of a given hand separately.