So we know a standard deck has 52 cards and it's asking the number of possible 5 card hands we could get from it without replacement which means without putting them back in the deck.
I thought about the usual formula of $52C5$ then multiplying it to $47C5$ => (47= 52-5), and so forth up until $7C5$ but I'm not sure that's the way.
If I could get a hint or an idea of if I'm going in the right way, that would really help me thanks!
Regards,
Assuming one hand is drawn
The question is equivalent to asking how many ways we can select $5$ objects from $52$ objects. As there is no $2$nd hand that one draws from the deck, the condition of replacement is redundant as the process already terminates after picking a deck of $5$ cards. Furthermore, all the cards in the deck are distinct. Hence, the answer simply is $52 \choose 5$.
Assuming ten hands are drawn
The condition of replacement makes a difference if ten hands are drawn consecutively, as the size of the deck shrinks by $5$ after every choice. Here, again, we must consider $2$ situations:
Assuming order of selection matters
In this case, your answer is correct, i.e.
$$\prod_{k=0}^9 {7 + 5k \choose 5}$$
Assuming order of selection does not matter
There are $10!$ ways to draw $10$ equivalent hands. Hence, in this case, the answer is
$$\frac{1}{10!}\cdot\prod_{k=0}^9 {7 + 5k \choose 5}$$