I just want to add a proof that may be given independently:
Suppose we are given $k$ mutually orthogonal vectors in ${\mathbb{R}}^n\backslash\{0\}$; in matrix form this is a $k\times n$ matrix  $A=[a_{i,j}]$. 
A $k$ choice of $k$ columns indices from $n$ is denoted by $\sigma_i$.

 The minor $M_{k,\sigma_i}$ is the minor formed by the columns with indices in $\sigma_i$ and rows $1$ to $k$.

The minor $M_{k-1,\sigma_i(\tilde{j})}$ is the minor formed by the columns with indices in $\sigma_i$ except $j$ ($j\in\sigma_i$) and rows $2$ to $k$.

For $n\ge 2$ fixed we want to prove $$(1) \quad \sum_{i=1}^{\binom{n}{k}}(M_{k,\sigma_i})^2=\prod_{i=1}^k(\sum_{j=1}^n a_{i,j}^2)$$
The cases $k=1$ and $k=2$ are either easy or trivial, suppose this is true for $k=r$, we want to prove it for $k=r+1$.

Expand the left hand side of $(1)$ with respect to the first row $$\begin{aligned}(2)\quad \sum_{i=1}^{\binom{n}{k}}(M_{k,\sigma_i})^2&=\sum_{i=1}^{\binom{n}{k}} \big(\sum_{j\in\sigma_i}a_{[1,j]}M_{k-1,\sigma_i(\tilde{j})} \big)^2\\&=\sum_{i=1}^{\binom{n}{k}}\big(\sum_{j\in\sigma_i}(a_{[1,j]}M_{k-1,\sigma_i(\tilde{j})})^2+2\sum_{\,\,j\neq j'\\j\&j'\in\sigma_i}a_{[1,j]}a_{[1,j']}M_{k-1,\sigma_i(\tilde{j})}M_{k-1,\sigma_i(\tilde{j'})}\big)\end{aligned}$$
where $a_{[1,j]}$ is the entry $a_{1,j}$ with the corresponding sign in $M_{k,\sigma_i}$

Set $M_{k-1,s_i}$ as the minor formed by rows $2$ to $k$ and columns indices in $s_i$ ($|s_i|=k-1$).
The idea is to make the remaining $M_{k-1,s_i}$ appear in $(2)$  so that we get $$(3)\quad(\sum_{j=1}^n a_{1,j}^2)(\sum_{i=1}^{\binom{n}{k-1}}M_{k-1,s_i})
$$ and from the induction hypothesis we can conclude.

Take the system on $k-1$ equations $$(4)\quad\sum_{j=1}^n a_{1,j}a_{h,j}=0\quad h=2,\ldots k$$

A choice of $k-2$ column indices is denoted by $t_i$. The set $t_i(l) $ ($l\notin t_i$) is the set $t_i$ with index $l$ added so $t_j(l)=s_i $ for some $i,j$. 
Multiply each equation in the system by a $k-2$ minor $r_h$ from columns indices in $t_i$ so that $$(5) \quad\sum_{h=1}^{k-1}r_h\sum_{j=1}^n a_{1,j}a_{h,j}=\sum_{l,l\notin t_i}\epsilon_l a_{1,l}M_{k-1,t_i(l)}=0$$ where ,$\epsilon_l=\pm 1$. 
Squaring the last equation for every choice $t_i$ and summing up we see that the number of product terms (with factor $2$) is $\binom{n}{k-2}\times \binom{n-k+2}{2}$ and this is equal to the product terms in $(2)$ or $\binom{n}{k}\times \binom{k}{2}$.

Finally notice that for every $t_i$, $l\neq l'$ and $l\&l'\notin t_i$ we have $$
\epsilon_l'\epsilon_l a_{1,l}a_{1,l'}M_{k-1,t_i(l)}M_{k-1,t_i(l')}=-a_{[1,l]}a_{[1,l']}M_{k-1,\sigma_i(\tilde{l})}M_{k-1,\sigma_i(\tilde{l'})}$$ with $t_i(l)=\sigma_i(\tilde{l'}),t_i(l')=\sigma_i(\tilde{l})$.

Replacing the product terms by the square ones in $(2)$ we obtain $(3)$.