Show that $\sqrt{\det J^T J}$ is equal to the 2-norm of the $\binom{m}{n}$ square sub-determinants of $J\in{\bf R}^{m\times n}.$
This is something I came across after a considerable amount of grief trying to figure out how to do a change of variables for the integration of an $n$ dimensional solid embedded in ${\bf R}^m$.
Basically, every introductory vector calculus book talks about how we have
$$\int_VdV=\int_W|\det J_F(x)|dx_1\ldots dx_m,$$
when $F:W\subseteq{\bf R}^m\to{\bf R}^m$.
However when $F:W\subseteq{\bf R}^n\to{\bf R}^m$ for $n<m$, then it's not obvious what to do since the Jacobian is no longer square. After a lot of reflection and experimentation, I realized that the appropriate generalization is this:
$$\int_VdV=\int_W\sqrt{\det(J_F(x)^TJ_F(x))}dx_1\ldots dx_n.$$
Believe it or not, this generalization is not mentioned on the Wikipedia pages for the Jacobian, or for the CoV for Integration.
As it happens, $\sqrt{\det J^T J}$ has another, more geometric, interpretation, being equal to the 2-norm of the $\binom{m}{n}$ square sub-determinants of $J\in{\bf R}^{m\times n}.$
A proof of this fact would probably be enlightening, but I'm also interested in how this relates to differential forms, as I'm pretty sure differential forms are just an abstraction/algebraification of this equality.
As per the answer given in the comments, this is a consequence of the Cauchy-Binet Formula.