# Let $A\in M_{n \times k}(\mathbb{R})$. Show that $\det(A^TA)=\sum \det(B^2)$ where the sum runs through all $k \times k$ submatrices $B$ of $A$.

Let $$A\in M_{n \times k}(\mathbb{R})$$. Show that $$\begin{equation} \det(A^TA)=\sum \det(B^2) \;\;\;\;\;\;\;(*) \end{equation}$$ where the summation runs through all $$k \times k$$ submatrices $$B$$ of $$A$$ that result from deleting rows of $$A$$.

Proof attempt:
If $$n, then $$A^TA$$, a $$k \times k$$ matrix, has rank at most $$n. Thus, it is singular and has determinant 0. Also, since $$n, there are no $$k \times k$$ submatrices of $$A$$ to consider. Thus, both sides of (*) are 0.

If $$n=k$$ the only $$k \times k$$ submatrix of $$A$$ is itself so the right hand side of (*) becomes $$\det(A^2)=\det(A)\det(A)=\det(A^T)\det(A)$$ which is equal to the left hand side.

However, I cannot seem to figure out the case for when $$n>k$$. I would appreciate an approach or a reference for a solution.

this is the Cauchy-Binet formula in the special case $$B=A^T$$.