Exercise involving DFT The fourier matrix is a transformation matrix where each component is defined as $F_{ab} = \omega^{ab}$ where $\omega = e^{2\pi i/n}$. The indices of the matrix range from $0$ to $n-1$ (i.e. $a,b \in \{0, ..., n-1\}$)
As such we can write the Fourier transform of a complex vector $v$ as $\hat{v} = Fv$, which means that $$\hat{v}_f = \sum_{a \in \{0,...,n-1\}}{\omega^{af}v_a}$$
Assume that $n$ is a power of $2$. I need to prove that for all odd $c \in \{0,...,n-1\}$, every  $d \in \{0,...,n-1\}$ and every complex vectors $v$, if $w_b = v_{cb+d}$, then for all $f \in \{0,...,n-1\}$ it is the case that : $$\hat{w}_{cf} =  \omega^{-fd} \phantom{a} \hat{v}_f$$
I was able to prove it for $n=2$ and $n=4$, so I tried an inductive approach. This doesn't seem to be the best way to go and I am stuck at the inductive step and I don't think I can go any further which indicates that this isn't the right approach.
Note that I am not looking for a full solution, just looking for a hint.
 A: Hint:
You need to use the fact that $\ n\ $ is a power of $\ 2\ $.  I believe a key observation is that because $\ c\ $ is odd, it is a unit in the ring of integers mod $\ n=2^r\ $ (for some positive integer $\ r\ $), and so, as $\ k\ $ ranges from $\ 0\ $ to $\ 2^r-1\ $ in the sum
$$
\hat{\omega}_{cf}=\sum_{k=0}^{2^r-1}e^\frac{2\pi i kcf}{2^r}v_{ck+d\pmod{2^r}}\ ,\\
$$
the residues of $\ ck\ $ mod $\ 2^r\ $ will range over the same set of values, but in a different order.  For each $\ k\in\big\{0,1,\dots,$$2^r-1\big\}\ $ there is a (unique) integer $\ u_k\ $ with $\ 0\le$$ u_k\le $$2^r-1\ $ such that $\ ck=u_k + z_k2^r\ $ for some integer $\ z_k\ $, and $\ u\ $ will be a permutation on the set of integers $\ \big\{0,1,\dots,2^r-1\big\}\ $.  We then have
\begin{align}
\sum_{k=0}^{2^r-1}e^\frac{2\pi i kcf}{2^r}v_{ck+d\pmod{2^r}}&=\sum_{k=0}^{2^r-1}e^\frac{2\pi i \left(u_k+z_k2^r\right)f}{2^r}v_{u_k+z_k2^r+d\pmod{2^r}}\ ,\\
 &=\sum_{k=0}^{2^r-1}e^\frac{2\pi iu_kf}{2^r}v_{u_k+d\pmod{2^r}}\ ,\\
&=\sum_{t=0}^{2^r-1}e^\frac{2\pi itf}{2^r}v_{t+d\pmod{2^r}}\ ,
\end{align}
the last expression being obtained by rearranging the order of summation.
