Determinant of a skew-symmetric matrix

Computation. $\det A = \det(A^T) = \det(-A)$. Since $A$ is $n\times n$, $\det(-A)=(-1)^n\det A$. Because $n$ is odd, $(-1)^n=-1$, so
$$\det A = -\det A.$$
Hence $2\det A=0$, so $\det A=0$.

Conclusion. Every real skew-symmetric matrix of odd order is singular.

Remark. In even dimension, the Pfaffian gives $\det A=(\operatorname{Pf}(A))^2$, which may be nonzero (e.g. $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ has determinant $1$).