How many different ways can 64 players be paired? 
Suppose that there are 64 players in an arena. In how many ways can
  the players be paired up (i.e. 32 different games)?

I think the answer would be:
$$\frac{\prod_{n=0}^{31} {64-2n\choose 2}}{32!}$$
I am not sure if this is a correct expression though, can anyone confirm? If it is, are there any simpler expression than the one I provided?
 A: Your answer is correct, and it is not hard to see what reasoning led to it.  Ideally, the expression should be accompanied by a justification.
There are other expressions for the answer. 
For example, let us line up the people in alphabetical order, or in order of height. Alicia can choose her partner in $63$ ways. For each such way, the first person not yet partnered can choose her partner in $61$ ways. And for each such way, the first person not yet partnered can choose her partner in $59$ ways. And so on. This gives a count of $(63)(61)(59)\cdots(3)(1)$. If one prefers (I don't) one can write this as 
$$\prod_{n=1}^{32} (2n-1).$$
Or else line up the $64$ people. This can be done in $64!$ ways.Now partner the first person with the second, the third with the fourth, and so on. Unfortunately this multiple counts the number of pairings. 
There are two reasons for this: (i) Any choice of interchanges of $2k-1$ with $2k$, $k=1$ to $32$, gives the same partnerings. To adjust, divide by the number of such possible interchanges, which is $2^{32}$. (ii) Any permutation of the $32$ couples gives the same partnerings. So we must further divide by $32!$. This yields the expression
$$\frac{64!}{2^{32}\cdot 32!}.$$
