Solution
= Solution
Applying the transform twice gives
$$
\operatorname{QFT}_N^2|x\rangle
=\frac1N\sum_{y,z=0}^{N-1}\omega^{y(x+z)}|z\rangle.
$$
By the <finite geometric series>, the inner sum over $y$ is $N$ when $x+z\equiv0\pmod N$ and zero otherwise. Therefore
$$
\operatorname{QFT}_N^2|x\rangle=|-x\bmod N\rangle,
$$
which is the stated $|N-x\rangle$ with the label interpreted modulo $N$.