Let $A_1\dots A_p$ be real $k\times k$ matrices, and define $kp\times kp$ matrix $Az$ as \begin{bmatrix} A_1z & A_2z & A_3z &\dots &A_{p-1}z & A_pz \\ I_kz & 0 & 0 &\dots &0 & 0 \\ 0 & I_kz & 0 &\dots &0 & 0 \\ \vdots & \vdots & \vdots &\dots &\vdots &\vdots \\ 0 & 0 &0 &\dots &I_kz &0 \end{bmatrix} where $z$ is a scalar.
In the Lutkepohl's (2005) book it is stated that one can (easily) prove that $$\det(I_{kp} - Az) = \det(I_k - A_1z - \dots - A_pz^p)$$
So far I have $$ \det(I_{kp} - Az) = \det \begin{bmatrix} I_k - A_1z & -A_2z & -A_3z &\dots &-A_{p-1}z & -A_pz \\ -I_kz & I_k & 0 &\dots &0 & 0 \\ 0 & -I_kz & I_k &\dots &0 & 0 \\ \vdots & \vdots & \vdots &\dots &\vdots &\vdots \\ 0 & 0 &0 &\dots &-I_kz &I_k \end{bmatrix} $$
I guess that one should apply suitable property of the determinants but I have troubles in finding such.
In your determinant perform column operations in blocks, adding into the first block column the second block column times $z$, the third times $z^2$ and so on. That will give you $$ \det \begin{bmatrix} I_k - A_1z - \dots - A_pz^p & -A_2z & -A_3z &\dots &-A_{p-1}z & -A_pz \\ 0 & I_k & 0 &\dots &0 & 0 \\ 0 & -I_kz & I_k &\dots &0 & 0 \\ \vdots & \vdots & \vdots &\dots &\vdots &\vdots \\ 0 & 0 &0 &\dots &-I_kz &I_k \end{bmatrix} $$ which gives $$\det(I_k - A_1z - \dots - A_pz^p)(\det I_k)^{p-1}$$ as required.