 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 $y={\sum }_{j=0}^{n-1}{2}^j{y}_j$. Then $z^y={\prod }_j(z^{2^j}{)}^{y_j}$, and henceup to the convention for ordering the binary digits. This explicit tensor product of one-qubit states proves that $|\varphi ⟩$ is a product state.