Finite length truncated exponential sequence $\mathcal Z$-transform, zeros over circle explanation
The $z$-transform is

$$\frac{1}{z^{N-1}}\cdot\frac{z^N-a^{N}}{z-a}$$
The zeros will therefore be given by
$$\begin{align}
z^N&amp;=a^{N}\cdot1\\
&amp;=\left(a^{N}\right)e^{j2k\pi}\\
&amp;=\left(ae^{\frac{j2k\pi}{N}}\right)^{N},\,k=1,2,\cdots,N-1
\end{align}$$ 
Hence,
$$\boxed{z_k=ae^\frac{j2k\pi}{N},\,k=1,2,\cdots,N-1}$$