Solution
= Solution
The relation $z^{2^n}=1$ makes $z$ a <root of unity>, so $|z|=1$ and normalization gives $T=2^{n/2}$. Write the <binary expansion> $y=\sum_{j=0}^{n-1}2^jy_j$. Then $z^y=\prod_j(z^{2^j})^{y_j}$, and hence
$$
|\phi\rangle
=\bigotimes_{j=0}^{n-1}
\frac{|0\rangle+z^{2^j}|1\rangle}{\sqrt2},
$$
up to the convention for ordering the binary digits. This explicit <tensor product> of one-qubit states proves that $|\phi\rangle$ is a <product state>.