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*}