Fix a vertex $v$ in $\vec{K_{2n}}$. To create a perfect matching, this vertex must be in a single edge. There are $2n-1$ ways to choose a neighbor $w$ for $v$ and $2$ ways to orient this edge. Now if we look at the subgraph on the remaining vertices, we see that we want a 1-factor on a complete directed graph with $2n-2$ vertices. So we can continue this process, fix some vertex $u$ other than $v,w$. We have $2n-3$ ways to choose a neighbor for $u$ and $2$ ways to orient this edge...
Overall, the number of 1-factors is $$ (2n-1) * 2 * (2n-3) * 2 * (2n-5) * 2 \cdots 5 * 2 * 3 * 2 * 1 *2 = 2^n * (2n-1)!!. $$ The double factorial is a common notation for this sort of product. Note that $$ (2n-1)!! = \frac{(2n)!}{2n*(2n-2) *(2n-4) \cdots 4*2} = \frac{(2n)!}{2^n n!}. $$ So the number of 1-factors of $\vec{K_{2n}}$ is $$ \frac{(2n)!}{n!}. $$