9. (16 points) Let $n$ be a positive odd number, and the complex number $z$ satisfies $z^{2^{n}-1}=1$. Find $Z_{n}=\prod_{k=0}^{n-1}\left(z^{2^{k}}+\frac{1}{z^{2^{k}}}-1\right)$.

Sure, here is the translated text:

```
9. Since $n$ is a positive odd number, we have
$$
2^{n}-1 \equiv(-1)^{n}-1 \equiv-2(\bmod 3) \text {. }
$$

Thus, $z^{3} \neq 1$.
Therefore, $z+\frac{1}{z}+1 \neq 0$.
By the difference of squares formula, we get
$$
\begin{array}{l}
\left(z+\frac{1}{z}+1\right) Z_{n} \\
=\left(z+\frac{1}{z}+1\right) \prod_{k=0}^{n-1}\left(z^{2^{k}}+\frac{1}{z^{2^{k}}}-1\right) \\
=z^{2^{n}}+\frac{1}{z^{2^{n}}}+1=z+\frac{1}{z}+1 .
\end{array}
$$

Hence, $Z_{n}=1$.
```