Determinant of $ADA^\top$, where $D$ is diagonal and $A$ is wide

Question

Let $A\in\mathbb{R}^{M\times N}$ be full rank, with $M<N$, and let $D\in\mathbb{R}^{N\times N}$ be diagonal, with a strictly positive diagonal.

Is it possible to simplify $\det{ADA^\top}$? In particular, can $D$ be separated out (in some sense)?

There have been previous questions on the eigenvalues of such a product here and here, but the determinant may just be slightly easier.

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Unsucessful attempts:

1. I considered starting from the fact that the determinant must be the product of the non-zero Eigenvalues of $D^{\frac{1}{2}}A^\top A D^{\frac{1}{2}}$. But while the determinant of square matrices is multiplicatively separable, the Eigenvalues are not.

2. I considered taking the QR decomposition of $A^\top$, so $A = [R^\top,0] [Q_{\cdot 1},Q_{\cdot 2}]^\top=R^\top Q_{\cdot 1}^\top$, where $[Q_{\cdot 1},Q_{\cdot 2}] [Q_{\cdot 1},Q_{\cdot 2}]^\top = I = [Q_{\cdot 1},Q_{\cdot 2}]^\top [Q_{\cdot 1},Q_{\cdot 2}]$, and where $R$ is square, invertible and upper triangular. Then the determinant of the original expression is $\left(\det{R}\right)^2 \det{\left(Q_{\cdot 1}^\top D Q_{\cdot 1}\right)}$. But you are still left with an expression of the same form as the original. (The fact that $Q_{\cdot 1}^\top Q_{\cdot 1}=I$ does not appear to help.)