Proof for the solution of homogenous difference equation
If you substitute your second equation back into the first equation, you obtain

$$\begin{align}\sum_{k=0}^{N}a_k\sum_{m=1}^NA_mz_m^{n-k}&amp;=\sum_{m=1}^NA_mz_m^n\sum_{k=0}^Na_kz_m^{-k}\end{align}\tag{1}$$

If $z_m$ are zeros of the polynomial

$$A(z)=\sum_{k=0}^Na_kz^{-k}\tag{2}$$

then Eq. $(1)$ equals zero, as required for the homogenous solution.