An identity with Gram determinant Let $\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{m}$ be $m$ vectors in $n$-dimensional space. Their Gram determinant is defined by:
$$\Gamma=\left|\begin{array}{cccc}
\mathbf{v}_{1}^{2} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\
\left(\mathbf{v}_{2}\cdot\mathbf{v}_{1}\right) & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\
\cdots & \cdots & \cdots & \cdots\\
\left(\mathbf{v}_{m}\cdot\mathbf{v}_{1}\right) & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2}
\end{array}\right|$$
If $v_{ij}$ is $j$th component of $\mathbf{v}_{i}$, prove that
$$\Gamma=\sum\left|\begin{array}{cccc}
v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\
v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\
\cdots & \cdots & \cdots & \cdots\\
v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}}
\end{array}\right|^{2}$$
where the summation is extended over all integers $s_{1},s_{2},\cdots,s_{m}$  from 1  to $n$  with $s_{1}<s_{2}<\cdots<s_{m}$.
 A: Here's how to begin:
Entry $(i,j)$ equals $\mathbf{v}_{i}\cdot\mathbf{v}_{j} = \sum_{s=1}^n v_{is} v_{js}$. Here $s$ is just a dummy variable, so we don't need to call it $s$ in each entry; let's choose to call it $s_j$ in each of the sums in column number $j$. Then, since the determinant is linear in each column separately, we can pull the sums and the factors $v_{j s_{j}}$ outside. Like this for the first column:
$$
\Gamma=\left|\begin{array}{cccc}
\sum_{s_1} v_{1 s_{1}} v_{1 s_{1}} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\
\sum_{s_1} v_{2 s_{1}} v_{1 s_{1}} & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\
\cdots & \cdots & \cdots & \cdots\\
\sum_{s_1} v_{m s_{1}} v_{1 s_{1}} & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2}
\end{array}\right|
=
\sum_{s_1} v_{1 s_{1}}
\left|\begin{array}{cccc}
v_{1 s_{1}} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\
v_{2 s_{1}} & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\
\cdots & \cdots & \cdots & \cdots\\
v_{m s_{1}} & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2}
\end{array}\right|.
$$
And doing the same for each column:
$$
\Gamma = 
\sum_{s_1} \sum_{s_2} \dots \sum_{s_m} v_{1 s_{1}} v_{2 s_{2}} \dots v_{m s_{m}}
\left|\begin{array}{cccc}
v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\
v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\
\cdots & \cdots & \cdots & \cdots\\
v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}}
\end{array}\right|.
$$
If two indices $s_j$ coincide, this last determinant has two equal columns and therefore vanishes, so we need only sum over index sets $(s_1,\dots,s_n)$ with all numbers distinct.
Can you see how to continue from here?
A: The most elegant solution is probably to extend the given scalar product on your $m$-dimensional vector space $V$ to the exterior product $\Lambda ^m(V)$.
The recipe is that any orthonormal basis $e_1,...,e_n$ of $V$ yields an orthonormal basis $e_{s_1 }\wedge ...\wedge e_{s_m }$                             $(s_{1}<s_{2}<\cdots<s_{m})$ of $\Lambda ^m(V) $ .
As a consequence, if you write $v_i=\Sigma v_{ij}e_j$ you get $$v_1\wedge...\wedge v_m=\Sigma c_{s_1...s_m} e_{s_1 }\wedge ...\wedge e_{s_m }  \quad (+)$$ with  
$$c_{s_1...s_m}= \left|\begin{array}{cccc}
v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\
v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\
\cdots & \cdots & \cdots & \cdots\\
v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}}
\end{array}\right|     \quad \quad (*)$$
Then taking the square of the lengths of both sides of this equality $(+) $ you obtain
$$||v_1\wedge...\wedge v_m||^2=\Sigma |c_{s_1...s_m}|^2\quad (**)$$   
Now remember that in a $m$- dimensional vector space spanned by  the vectors $v_1,..., v_m$ (assumed independent: everything is trivial if these vectors are dependent ) we have the equality for  squared length $$||v_1\wedge...\wedge v_m||^2=\Gamma =\left|\begin{array}{cccc}
\mathbf{v}_{1}^{2} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\
\left(\mathbf{v}_{2}\cdot\mathbf{v}_{1}\right) & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\
\cdots & \cdots & \cdots & \cdots\\
\left(\mathbf{v}_{m}\cdot\mathbf{v}_{1}\right) & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2}
\end{array}\right|\quad \quad (***)$$ ( determinant of Gram matrix).
Replacing both sides of $(**)$
 by their values given in $(*)$  and $(***)$ proves the  required formula.
Edit
1) I forgot to mention that  formula $(*)$ can be thought of as a generalization of Pythagoras' theorem, with $||v_1\wedge...\wedge v_m||^2$ playing the role of the squared length  of the hypotenuse $v_1\wedge...\wedge v_m$.  
2) I feel that the euclidean structure inherited by $\Lambda ^mV \;\;$ from a euclidean structure on $V$ is not as well-known as it deserves.
 One of the rare treatments of this theme in the textbook literature is given in MacLane-Birkhoff's Algebra (cf. especially Theorem 18, page 557).  
