I have been trying to prove the well-known relation linking the Pfaffian and the determinant from their definitions in terms of the exterior algebra, but unfortunately I haven't been able to work it out myself, nor to find the proof elsewhere.
Here go the definitions:
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.
As for the determinant, we know that $f$ induces an $\bigwedge^{2n} f: \bigwedge^{2n} E \rightarrow \bigwedge^{2n} E$. Now, linear endomorphisms of one-dimensional vector spaces are just multiplication by a scalar. In our case, we define:
$$(\bigwedge^{2n} f) (e_1\wedge e_2 \wedge ... \wedge e_{2n})=det(f)e_1\wedge e_2 \wedge ... \wedge e_{2n}$$.
From this definitions one should be able to derive the well-known formula:
$$Pf(f)^2=det(f)$$
Any help would be much appreciated.