dividing a group 0f 10 men and women into couples of same sex 
There're 10 men and 10 women. We need to divide them into couples of same sex. 

The answer is: $$\left( \frac{10!}{5!*2^5} \right)^2$$ I don't understand why. 
I know that we can divide into couples where there's exactly 1 man and 1 woman in each couple in $10!$ ways (1). Now I would need to find the total possible number of couples regardless of sex $\left( \frac{20!}{10!*2^{10}} \right)$ (2). Then (2) - (1) should give the answer. But that doesn't give the same result.

You are right about the total possible number of couples regardless of sex.  The number $\frac{20!}{10!*2^{10}}$ includes every possibility: out of the 10 couples, they can be of the same gender, or some of them can be of 1 man and 1 woman.  The number $10!$ represents the case in which none of the couples are of the same sex, so you can subtract $10!$ from that to get the answer.
We first divide them into two groups of 10 persons by sex.  In each of these two groups,
we have $\binom{10}{2}$ ways to choose the first couple while leaving $8 \times 2$ persons behind, then we have $\binom{8}{2}$ ways to pick another couple while leaving $6 \times 2$ persons behind.  We continue this process until everyone has its partner.  Therefore, we have
$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$, but we need to divide this by $5!$ since we ask for the number of ways of dividing them into couples.  This gives the desired result for the group of 10 persons.  Squaring this will give the required answer.
\begin{align}
\frac{1}{5!}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} &= \frac{1}{5!} \frac{10\times9}{2}\,\frac{8\times7}{2}\,\dots\frac{2\times1}{2} \\
&= \frac{10!}{5!*2^{5}}
\end{align}
