Simple/Concise proof of Muir's Identity I am not a Math student and I am having trouble finding some small proof for the Muir's identity.
Even a slightly lengthy but easy to understand proof would be helpful.
Muir's Identity
$$\det(A)= (\operatorname{pf}(A))^2;$$
the identity is given in the first paragraph of the following link
http://en.wikipedia.org/wiki/Pfaffian
I am expecting a proof which uses minimal advanced mathematics.Any reference to a textbook or link would do.
I would be very grateful, if any of you could point me in that direction.
P.s- i have done all the googling required and i wasnt satisfied with their results,so dont post any results from google's 1st page
Thanks in Advance
 A: 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. 
