Finite length truncated exponential sequence $\mathcal Z$-transform, zeros over circle explanation I'm looking at an example on how to obtain the $\mathcal Z$-transform from a finite length truncated exponential sequence, namely: 
$$x[n] = 
\begin{cases}
a^N &amp;\text{for} &amp; 0 \leq n \leq N-1\\[2ex]
 0 &amp; \text{otherwise}
\end{cases}
$$

The $\mathcal Z$-transform from that sequence is:
$$
\frac {1}{z^{N-1}} \cdot \frac {z^{N} - a^N}{z - a}
$$

It can be seen that there are $N-1$ poles at the origin and $N-1$ zeros at $z = a$ (because the pole in $z=a$ is cancelled by a zero). My doubt is on the location of the zeros over the $z$-plane.

I know that the $N-1$ zeros at $z = a$ can be expressed as:

$z = a\cdot1\ \text{and}\ 1 = e^{j2\pi k}$ so:
$$
z_k = ae^{j2\pi k}\quad\text{with}\quad k = 1,2,\ldots,N-1
$$ 
Yet in the example I'm looking at, they say that the zeros are located at 
$$z_k = ae^{j\frac{2\pi k}{N}}\quad\text{with}\quad k = 1,2, \ldots, N-1$$
So they are spaced at $\frac{2\pi}{N}$ instead of being all at the same place and I don't understand where that $N$ came from.
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,
$$\boxed{z_k=ae^\frac{j2k\pi}{N},\,k=1,2,\cdots,N-1}$$