Geometrcal proof of "Pfaffian squared is determinant" Let $(V,g)$ be an oriented Euclidian vector space of dimension $2n$, with a non-singular operator $A\in End(V)$ anti-symmetric, then $w_A(u,v)=g(u,Av)$ defines a linear symplecitc form. Define its Pfaffian by 
$\frac{w_A^n}{n!}=Pf(A)\mathrm{vol}$, where $\mathrm{vol}$ denotes the volume form of $V$. 
You can pull back the relation $\frac{w_A^n}{n!}=Pf(A)\mathrm{vol}$ by a $B\in End(V)$. You will get
$$Pf(B^TAB)=Pf(A)\det(B). $$
Use the normal form theorem in linear symplectic geometry, we may assume $V=\mathbb{R}^{2n}$ and $(w_0,g_0,J_0)$ denotes the standard compatible triple and there's $P:(V,w_A)\to (V,w_0)$ symplectomorphism.
$$w_A(u,v)=g_0(u,Av)=P^*w_0(u,v)=w_0(Pu,Pv)=g_0(J_0Pu,Pv)=g_0(P^TJ_0Pu,v)=g_0(-Au,v).$$
So, $A=P^TJ_0^{-1}P$, and $Pf(A)^2=\det(A)$.
But this proof uses the normal form theorem. Is there a more geometrical one without citing this theorem?
 A: Here is another idea of how to prove it geometrically, but still using a basis theorem.
$\newcommand{\innprod}[1]{\left\langle #1 \right\rangle}$

Theorem: Let $V$ be a finite-dimensional real vector space, equipped with an inner product $\innprod{-, -} \colon V \times V \to \mathbb{R}$ and a symplectic form $\omega \colon V \times V \to \mathbb{R}$. Then there exists an orthonormal basis $e_1, \ldots, e_n, f_1, \ldots, f_n$ and scalars $\lambda_1 \geq \cdots \geq \lambda_n > 0$ such that the pairing $\omega$ on the basis is given by
$$ \omega(e_i, e_j) = \omega(f_i, f_j) = 0, \quad \omega(e_i, f_j) = \lambda_i.$$
Furthermore, the set of scalars is canonically determined, and when $A_\omega \colon V \to V$ is the skew-symmetric isomorphism associated to $\omega$, it acts on the basis by $A_\omega e_i = \lambda_i\;\!\;\! f_i$ and $A_\omega f_i = - \lambda_i e_i$.

Perhaps you have come across this theorem, or perhaps in a slightly different form. It is not difficult to prove, I can outline the steps if you like.
But now it is simple to check that $\det A = \lambda_1^2 \cdots \lambda_n^2$, and that
$$ \omega^n(e_1, \ldots, e_n, f_1, \ldots, f_n) = n! \, \lambda_1 \ldots \lambda_n \operatorname{vol}(e_1, \ldots, e_n, f_1, \ldots, f_n)$$
hence $\operatorname{Pf}(\omega) = \pm \lambda_1 \ldots \lambda_n$ and indeed we have $\operatorname{Pf}(\omega)^2 = \det(A)$.
I would count this proof as very geometric, since the actual definition of Pfaffian (of a linear operator) depends on understanding the interaction between the orthogonal structure $\innprod{-, -}$ and the induced symplectic structure $\omega(-, -)$.
A: Here is an idea for a proof without the normal form theorem, but of unknown geometric interpretation.
For a skew-symmetric $2n \times 2n$ matrix $A$
consider the following simultaneoues row and column operations.
$$
C :=
\begin{pmatrix}
A & 0 \\
0 & A
\end{pmatrix}
\rightsquigarrow
\begin{pmatrix}
A & iA \\
iA & 0
\end{pmatrix}
\rightsquigarrow
\begin{pmatrix}
0 & iA \\
iA & 0
\end{pmatrix}
= B^T \Omega B
$$
where
$\Omega =
\begin{pmatrix}
0 & I_{2n} \\
-I_{2n} & 0
\end{pmatrix}$
and
$B :=
\begin{pmatrix}
iA & 0 \\
0 & I_{2n}
\end{pmatrix}$
which is again skew-symmetric.
Or in other words
$$
C = T^T B^T \Omega B T
\qquad\text{with}\qquad
T = 
\begin{pmatrix}
I_{2n} & 0 \\
-\frac{i}{2} I_{2n} & I_{2n}
\end{pmatrix}
\begin{pmatrix}
I_{2n} & -i I_{2n} \\
0 & I_{2n}
\end{pmatrix}
$$
Then the transformation property of the Pfaffian under base change gives
$$
\newcommand{Pf}{\operatorname{Pf}}
\Pf(A)^2 
= \Pf(C)
%= \Pf(T^T B^T \Omega B T)
= \Pf(\Omega) \det(B) \det(T)
= \det(A)
$$
where we used
$\Pf(\Omega) = (-1)^\frac{2n(2n-1)}{2} = (-1)^n$
and
$\det(B) = \det(iA) = (-1)^n \det(A)$.
