For $A,B,C,D$ matrices of appropriate sizes, you have the formulae
$$\det(M)= \det(A-BD^{-1}C)\det(D)$$ where
$$M = \begin{pmatrix}
A & B\\
C & D
\end{pmatrix}.$$
Which gives in your particular case that
$$\det(M) = \det(I_k+A^TA)\det(-I_l).$$
We're left to prove that $\det(I_k+A^TA) \neq 0$, i.e. that $-1$ is not an eigenvalue of $A^TA$. And indeed if that was the case, we would be able to find $X \neq 0$ such that
$$\Vert AX \Vert^2 = X^T A^T A X = -X^TX = -\Vert X \Vert^2$$ and that can't be as the left side of the equality is non negative while the right side is strictly negative.
Note: following the answer of "levap", I made here the assumption that $\mathbb F$ is $\mathbb R$ or a formally real field.