This is admittedly not very hard even without getting the trick, but it's super-easy with the trick. Hopefully it isn't too easy to be interesting (or even amusing):

Problem. Let $v\_1,\ldots,v\_n$ be vectors in $\mathbb{R}^m$, and let $V$ be the $m\times n$ matrix whose columns are $v\_1,\ldots,v\_n$. Show that the $n$-dimensional volume of the $n$-dimensional parallelepiped in $\mathbb{R}^m$ determined by $v\_1,\ldots,v\_n$ is $\sqrt{\det(V^TV)}$.

We should probably take as given that the determinant of a square matrix is the signed volume of the parallelepiped determined by its columns (or by its rows); I would consider justifying that to be a separate problem (for which I don't know any simple tricks).

This is admittedly not very hard even without getting the trick, but it's super-easy with the trick. Hopefully it isn't too easy to be interesting (or even amusing):

Problem. Let $v_1,\ldots,v_n$ be vectors in $\mathbb{R}^m$, and let $V$ be the $m\times n$ matrix whose columns are $v_1,\ldots,v_n$. Show that the $n$-dimensional volume of the $n$-dimensional parallelepiped in $\mathbb{R}^m$ determined by $v_1,\ldots,v_n$ is $\sqrt{\det(V^TV)}$.

We should probably take as given that the determinant of a square matrix is the signed volume of the parallelepiped determined by its columns (or by its rows); I would consider justifying that to be a separate problem (for which I don't know any simple tricks).