I have worked with the definition of the Pfaffian given in terms of the exterior algebra:
Let $E$ be an even dimensional vector space ($dimE=2n$) equipped with a Euclidean metric $g$. Let $f:E \rightarrow E$ be an antisymmetric endomorphism. Then $f$ induces a form $\omega \in \bigwedge^2 E$ given by $\omega(w, v)=g(f(w), v)$. This is equivalent to defining $\omega = \sum_{i<j}a_{ij} e_i\wedge e_j $, where $(a_{ij})$ is the matrix of $f$ with respect to an orthonormal basis of $E$.
Now, if we take $\omega^n$ -this means $\omega \wedge \omega ...$, $n$ times-, we realise $\omega^n \in \bigwedge^{2n} E$, which is one-dimensional, so that we must have:
$$ \omega^n= Pe_1\wedge e_2 \wedge ... \wedge e_{2n} $$ This constant $P$ is called the Pfaffian.
The usual definition of the Pfaffian, however, is given in terms of permutations, via:
$$ \displaystyle{ \operatorname{pf}(A) = \frac{1}{2^n n!}\sum_{\sigma\in S_{2n}}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}a_{\sigma(2i-1),\sigma(2i)} }, $$
for a skew-symmetric matrix A of even dimension.
I would like to prove that the two definitions are equivalent, but unfortunately I haven't managed to.