Inverse $\mathcal Z$-transform of system with an 8th order pole
Applying the geometric series directly

$$
H(z) = \frac{1}{1-\alpha z^{-8}} = \sum_{n=0}^{\infty}(\alpha z^{-8})^n = \sum_{n=0}^{\infty}\alpha^n z^{-8n}
$$
If $h_n = \mathcal Z^{-1}\{H(z)\}$, then
$$
h_n = \begin{cases}
a^{n/8} \; , \;  \text{n is multiple of 8} \\
0 \; , \; \text{otherwise}
\end{cases}
$$