In how many ways can we pair ourselves?

We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is

$$\dfrac{(2n)!}{2^nn!}.$$

Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.

Denote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\qquad P_{2n}=(2n-1)\>P_{2n-2}\qquad(n>1)\ ,$$ which immediately leads to you "wild guess".

$$\frac1{n!}\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^nn!}$$

Pick out $2$ out of all $2n$ to form a pair.

After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera.

Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this.