**Problem: We have matrix $A\in M\_{n-1\times n}(\mathbb Z)$ so that the sum of elements in each row is zero. Prove that $\det(AA^T)=nk^2$, where $k\in \mathbb Z$.**

What have I considered so far:

1. First I thought, since sum of all elements in each row is zero, zero is eigenvalue of $A$ but $A\in M\_{(n-1)\times n}(\mathbb Z)$ confused me.
2. I see that $AA^T$ will be $(n-1)$ by $(n-1)$, so I tried calculating $AA^T$ but I failed to see any connection.

I did some research and found [this](https://math.stackexchange.com/questions/583938/do-non-square-matrices-have-eigenvalues) question but we had no mention of it, which makes me believe I am on the wrong track.

Thank you all for your help.