We have been told in class that the Pfaffian of $2n$ by $2n$ skew-symmetric matrix $A$ is defined as: $$Pf(A)=\frac{1}{2^nn!}\sum_{\sigma\in S_{2n}}A_{\sigma(1)\sigma(2)}\cdots A_{\sigma(2n-1)\sigma(2n)}.$$
And with this definition, I can show that $Pf(B^tAB)=\det(B)Pf(A)$ by brute force computations. However, I am stumped at showing the fact that $Pf(A)^2=\det(A)$.
Since every skew-symmetric matrix of size $2n$ can be orthogonally similar to a direct sum of $2$ by $2$ skew-symmetric in the form $$M_i=\begin{bmatrix}0&m_i\\-m_i&0\end{bmatrix}$$ and I've proved that the $Pf(M_i)=m_i$. If I am able to show that $$Pf\left(\bigoplus_i M_i\right)=\prod_{i} Pf(M_i).$$ Then this can be done easily, but this is where I am stuck.
It's enough to show that $\texttt{Pf}(M_1 \oplus M_2)=\texttt{Pf}(M_1)\cdot \texttt{Pf}(M_1)$. So suppose $M_1$ has dimensions $2a_1$ and $M_2$ $2a_2 = 2a-2a_1$.
Consider $\texttt{Pf}(M_1\oplus M_2)=\frac{1}{2^aa!}\sum _{\sigma \in S_{2a}}\prod _{i\in [a]}(M_1\oplus M_2)_{\sigma(2i-1)\sigma(2i)},$ and notice that if $\sigma(2i-1)\in [a_1]$ and $\sigma(2i)\not \in [a_1]$ or viceversa, then $(M_1\oplus M_2)_{\sigma(2i-1)\sigma(2i)}=0$ because its going to be in a part of a matrix with just $0's$, this implies that we can decompose $\sigma$ into two permutations and the positions in which the first permutation is. That is, consider $\varphi : S_{2a}\longrightarrow S_{a_1}\times S_{a_2}\times \binom{[a]}{a_1}$ where $\varphi (\sigma)=(\sigma _1,\sigma _2,X)$. Here $X$ is going to be the set of $i\in [a]$ such that $\sigma(2i-1),\sigma (2i)\in [a_1]$. If $X = \{x_1<x_2,\cdots<x_{a_1}\},$ then $\sigma_1 = \sigma(2x_1-1)\sigma(2x_1)\sigma(2x_2-1)\sigma(2x_2)\cdots\sigma(2x_{a_1}-1)\sigma(2x_{a_1})$ and if $[a]\setminus X = \{y_1<y_2<\cdots <y_{a_2}\},$ then $\sigma_2 = (\sigma(2y_1-1)-2a_1)(\sigma(2y_1)-2a_1)(\sigma(2y_2-1)-2a_1)(\sigma(2y_2)-2a_1)\cdots(\sigma(2y_{a_2}-1)-2a_1)(\sigma(2y_{a_2})-2a_1))$
Turns out that $\varphi$ is a bijection, because clearly you can inverse the process, just put the elements of $\sigma_1$ on the indices given by $X$, and so $$\texttt{Pf}(M_1\oplus M_2)=\frac{1}{2^aa!}\sum _{\substack{\sigma \in S_{2a}\\f(\sigma)=(\sigma_1,\sigma _2,X)\\(\sigma_1,\sigma _2,A) \in S_{a_1}\times S_{a_2}\times \binom{[a]}{a_1}}}\prod _{i\in [a]}(M_1\oplus M_2)_{\sigma(2i-1)\sigma(2i)},$$ using commutativity of the product, one can see that $X$ is irrelevant for the product because one only needs consecutive elements to be in the same set(either $[a_1]$ or $[a]\setminus [a_1]$), and so $$=\frac{1}{2^aa!}\sum _{X\in \binom{[a]}{a_1}} \sum _{\substack{(\sigma_1,\sigma _2) \in S_{a_1}\times S_{a_2}}}\prod _{i\in [a_1]}(M_1)_{\sigma _1(2i-1)\sigma _1(2i)}\prod _{i\in [a_2]}(M_2)_{\sigma _2(2i-1)\sigma _2(2i)}$$ $$=\frac{1}{2^aa!}\binom{a}{a_1} \sum _{\substack{(\sigma_1,\sigma _2) \in S_{a_1}\times S_{a_2}}}\prod _{i\in [a_1]}(M_1)_{\sigma _1(2i-1)\sigma _1(2i)}\prod _{i\in [a_2]}(M_2)_{\sigma _2(2i-1)\sigma _2(2i)}$$ $$=\frac{1}{2^aa_1!(a-a_1)!}\left (\sum _{\sigma_1 \in S_{2a_1}}\prod _{i\in [a_1]}(M_1)_{\sigma _1(2i-1)\sigma _1(2i)}\right )\left (\sum _{\sigma_2 \in S_{2(a-a_1)}}\prod _{i\in [a_2]}(M_2)_{\sigma _2(2i-1)\sigma _2(2i)}\right)$$ $$=\texttt{Pf}(M_1)\cdot \texttt{Pf}(M_2)$$
BtW, this is Cayley's Theorem and there is a very known bijective proof, you can check it out in Chapter 5 of Martin Aigner's "A course in enumeration".