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} 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).