Trace of Product of Kronecker Products Start Wearing Purple answered this question for me. I am now asking a question about their answer as I can make the question general enough to be useful on it's own.
For $1\times 2$ matrices $\alpha,\,\beta\in M_{1\times 2}(\mathbb{C})$, SWP uses the following equality
$$\operatorname{Tr}\left[(\overline{\alpha}\otimes\beta^*)^k(\alpha^T\otimes \beta)^k\right]=(\bar{\alpha}\cdot \beta^*)^{k-1}(\overline{\alpha}\cdot \alpha^T)(\beta\cdot \alpha^T)^{k-1}(\beta\cdot \beta^*).\,\,\,\,\,\,(\star)$$
where the dots are scalar products/matrix multiplications.
To be honest I don't really have a clue how to deal with these having never used Kronecker products before. I have used the result in my work but do not yet understand the above calculation.

I am looking for somebody to explain this calculation $(\star)$ to me or perhaps
  send me to an appropriate online reference.

There is a chance that the exact nature of $\alpha$ and $\beta$ make this calculation true:
$$\alpha:=(1\,\,\,\,\,\zeta^{-v})\,\,\,\text{ and }\,\,\,\beta:=(\zeta^{-v}\,\,\,\,\, \zeta^v),$$
for $\zeta:=e^{2\pi i/n}$.
Context: I used this result to help bound the number of steps required for a random walk on the Sekine quantum groups to converge to the Haar state. More background here. 
 A: We will need a few things:
\begin{align}
\text{ conjugate-rule:}&&&  (A\otimes B )^*=  A^* \otimes B^* &\text{note they don't change places!}
\\
\text{ mixed-product-rule:}&&&  (A\otimes B )(C \otimes D) = (AC)\otimes (BD)
&\text{if matrices have correct dim}
\\
\text{ trace-rule:}&&&  {\tt tr}(A \otimes B) =  {\tt tr}(A) {\tt tr}(B)
\\
\text{ trace-product:}&&&  {\tt tr}(A^* B) =  {\tt tr}(A B^*)
\\
\text{ trace-linearity:}&&&  {\tt tr}(\lambda A) = \lambda {\tt tr}(a) &\text{for scalar $\lambda$}
\end{align}
Now note that since $\alpha$ and $\beta$ are $1\times 2$,
$${\tt (I)}\qquad
(\alpha^T\otimes \beta)^*  (\alpha^T\otimes \beta) = (\bar\alpha\otimes\beta^*)(\alpha^T\otimes \beta) =  (\underbrace{\bar\alpha\alpha^T}_{1\times 1} \otimes \underbrace{\beta^*\beta}_{2\times 2}) = (\bar\alpha\alpha^T)(I_1 \otimes \beta^*\beta)
$$
And now we on the one hand: 
$${\tt (II)}\qquad
(\alpha^T\otimes \beta)^* (I_1 \otimes \beta^*\beta) = (\bar\alpha\otimes \beta^*) (\beta^*\beta \otimes I_1) = ((\underbrace{\bar\alpha \beta^*}_{1 \times 1})\beta \otimes \beta^*) = (\bar\alpha \beta^*)(\beta \otimes \beta^*)
$$
and on the other
$$
{\tt (III)}\qquad
(\beta \otimes \beta^*)(\alpha^T\otimes \beta) = (\underbrace{\beta\alpha^T}_{1\times 1}\otimes \beta^*\beta) = (\beta\alpha^T)(I_1 \otimes \beta^*\beta)
$$
So Combining (I), (II) and (III) we find that by sort of 'ping-ponging' the middle Kroner-product back and forth that we can extract more and more scalar terms by reducing the powers:
\begin{align}
\big((\alpha^T\otimes \beta)^*\big)^k  (\alpha^T\otimes \beta)^k 
&=
(\bar\alpha\alpha^T) \big((\alpha^T\otimes \beta)^*\big)^{k-1} (I_1 \otimes \beta^*\beta)(\alpha^T\otimes \beta)^k 
\\&=
(\bar\alpha\alpha^T)(\bar\alpha \beta^*)^{k-1}(\beta\alpha^T)^{k-1}(I_1 \otimes \beta^*\beta)
\\&=
(\bar\alpha\alpha^T)(\bar\alpha \beta^*)^{k-1}(\beta\alpha^T)^{k-1}(\beta^* \beta)
\end{align}
From which your identity can be immediately deduced by the trace linearity and product rule:
$$ {\tt tr}(\underbrace{\beta^*\beta}_{2x2}) =  {\tt tr}(\underbrace{\beta\beta^*}_{1 \times 1}) = \beta\beta^*$$
Note that since $\beta$ is a row-vector holds: $(I_1 \otimes \beta^*\beta) = (\beta^*\beta \otimes I_1) = \beta^*\beta =(\beta^*\otimes \beta )$. In fact all of the above calculations immediately generalize to the case of $1\times M$ vectors.
If you want to I can also post an answer how to obtain the equality 
$${\tt tr}((A^*)^k A^k) = 4\left(1+\zeta^v\right)^{2k-1}\left(1+\zeta^{-v}\right)^{2k-1}$$ from your original post without the need of using Kronecker products.
