Let $A = [a_{i,j}]_{i,j} \in M_{2n}(\mathbb{R})$ be a skew-symmetric matrix such that $a_{i,j} \neq 0$, $\forall i \neq j$. I am trying to see if it's true that $\det A \neq 0$.
I tried working from the definition of $\det A$; the only useful observations I could see were that the sum over permutations reduces to a sum over permutations without fixed points, and that for a permutation $\sigma$ we have by skew-symmetry $$ \prod\limits_{i=1}^{2n} a_{i, \sigma(i)} = \prod_{i=1}^{2n} a_{i, \sigma^{-1}(i)} $$
But I can't see how and whether these are useful. I also know that the determinant is the square of the Pfaffian, but I can't see how to use this fact either.
Here it is best to consider an example. Then we can see that the Pfaffian polynomial can get zero for certain choices. So let $n=2$ and consider the following skew-symmetric matrix $$ A=\begin{pmatrix} 0 & 3 & 1 & -1 \\ -3 & 0 & 2 & 1 \\ -1 & -2 & 0 & 1 \\ 1 & -1 & -1 & 0 \end{pmatrix} $$ Using the determinant formula for skew-symmetric matrices of size $4$, we see that $\det(A)=0$. For a $4\times 4$ skew-symmetric matrix the general formula for its determinant is $$ (af-be+dc)^2 $$ for the matrix $A$ in $a,b,c,d,e,f$ from your wikipedia link.