This answer does not show the explicit form of $\textrm{pf}(A)$ but it proves that such a form must exist as a polynomial in the entries of $A$.
Let $A$ be a generic skew-symmetric $n \times n$ matrix with indeterminate entries $A_{i j}$ on row $i$ column $j$ for $0 \leq i < j \leq n$. I will prove by induction in $n$ that $\det(A)$ is the square of a polynomial in the indeterminates $A_{i j}$. If $n = 2$ then
$$ \det \begin{pmatrix} 0& A_{1 2} \\ -A_{1 2}& 0
\end{pmatrix} = A_{1 2}^2 $$
is a square polynomial.
Let $$
B = \begin{pmatrix}1 \\
& 1 \\
& -A_{1 3} & A_{1 2} \\
& -A_{1 4} & & A_{1 2} \\
& \vdots & & & \ddots \\
& -A_{1 n} & & & & A_{1 2}
\end{pmatrix}
$$
where all unlisted entries are equal to zero. Then the product
$$ C = B\,A\,B^{T} $$
is skew-symmetric and takes the form
$$
C = \begin{pmatrix} 0& A_{1 2} & 0 & \dotsc & 0\\
-A_{1 2} & 0 & \ast & \dotsc & \ast\\
0 & \ast & \ddots & \ddots & \vdots\\
\vdots & \vdots & \ddots & & \ast \\
0 & \ast & \dotsc & \ast & 0
\end{pmatrix}
$$
where each asterisk denotes some polynomial in the indeterminates. Let $A'$ be the bottom right $(n-2) \times (n-2)$ skew-symmetric sub-matrix of $C$. By induction $\det(A')$ is the square of a polynomial. From the explicit form of $C$ it follows that $$ A_{1 2}^{2n-4} \det(A) = \det(B \, A \, B^T) = \det(C) = A_{1 2}^2 \det(A') $$ or $$\det(A) = A_{1 2}^{6-2n} \det(A').$$ Now the right hand side must be a polynomial (because $\det(A)$ is) and since it is also a square we are done.
