Note that the inverse of the leading principal $(n-1)p\times(n-1)p$ submatrix is given by $ \begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix}^{-1} =\begin{pmatrix} \lambda^{-1} I_p & \lambda^{-2}I_p & \cdots & \lambda^{1-n}\, I_p\\ & \ddots &\ddots &\vdots\\ & & \ddots & \lambda^{-2}I_p \\ & & & \lambda^{-1} I_p \end{pmatrix} $ Therefore, using Schur complement, your determinant evaluates to \begin{align*} &\det\begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix} \times \\ &\det\left[(\lambda I_p +A_{n-1}) - (A_0, A_1, \ldots, A_{n-2}) \begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix}^{-1} \begin{pmatrix}0\\ \vdots\\ 0\\-I_p\end{pmatrix}\right]\\ =&\lambda^{n-1} \det\left[(\lambda I_p +A_{n-1}) + (A_0, A_1, \ldots, A_{n-2}) \begin{pmatrix}\lambda^{1-n}\,I_p\\ \vdots\\ \lambda^{-2}\,I_p\\ \lambda^{-1}I_p\end{pmatrix}\right]\\ =&\det( A_0+ A_1 \lambda +...+ A_{n-1}{\lambda}^{n-1}+ I_p {\lambda}^n). \end{align*}