How to derive the determinant of the variance of a negative multinomial distribution? The probability mass function of the negative multinomial distribution is:
\begin{align*}
\mathbb{P}(\boldsymbol{\rm{X}}=\boldsymbol{\rm{x}}|\mathbf{p})=\frac{\Gamma\left(x_0+\sum_{i=1}^{m}x_{i}\right)}{\Gamma\left(x_0\right)}p_0^{x_0} \prod_{i=1}^{m}\frac{p_{i}^{x_i}}{x_i!},
\end{align*}
where $\mathbf{p} \equiv (p_1,...,p_m)\in\mathbb R_+^m$ and $p_0 \equiv 1-\sum_{i=1}^m p_i$.  Taking $x_0$ to be fixed as the number of successful draws from the "first" group, the mean vector and variance matrix of the distribution of the remaining $m$ values are given respectively by:
\begin{align*}
 \boldsymbol{\rm{\mu}}=\frac{x_0}{p_{0}}\boldsymbol{\rm{p}}
\quad \quad \quad \quad \quad
{\boldsymbol {\Sigma }}=\frac{x_0}{p_{0}^{2}}\boldsymbol{\rm{p}}\boldsymbol{\rm{p'}}+\frac{x_0}{p_{0}}{\rm{diag}(\boldsymbol{\rm{p}})}.
\end{align*}
I want to show that the determinant of the variance matrix is:
\begin{align*}
 {\displaystyle |{\boldsymbol {\Sigma }}|={\frac {1}{p_{0}}}\prod _{i=1}^{m}{\mu _{i}}},
\end{align*}
where the values in the product are elements of the mean vector.  How can I derive this result?

As a preliminary step, we note the result of the following quadratic form:
$$\mathbf{p}' \text{diag} (\mathbf{p})^{-1} \mathbf{p}
= \mathbf{p}' \mathbf{1}
= \sum_{i=1}^m p_i
= 1-p_0.$$
Now, using the matrix determinant lemma, we have:
$$\begin{align}
|\boldsymbol{\Sigma}|
&= \bigg| \frac{x_0}{p_{0}^{2}} \mathbf{p} \mathbf{p}' + \frac{x_0}{p_{0}}\text{diag} (\mathbf{p}) \bigg| \\[6pt]
&= \bigg( 1 + \frac{\mathbf{p}' \text{diag} ( \mathbf{p})^{-1} \mathbf{p}}{p_0} \bigg) \bigg| \frac{x_0}{p_{0}} \text{diag} (\mathbf{p}) \bigg| \\[6pt]
&= \bigg( 1 + \frac{1-p_0}{p_0} \bigg) \prod_{i=1}^m \frac{x_0}{p_0} p_i \\[6pt]
&= \frac{1}{p_0} \prod_{i=1}^m \frac{x_0}{p_0} p_i \\[6pt]
&= \frac{1}{p_0} \prod_{i=1}^m \mu_i. \\[6pt]
\end{align}$$
As you can see, the result you are interested in is a direct result of the matrix determinant lemma; if you are interested in a proof of the latter you can find it here.
