Is this proof correct? (Notation: $\zeta_n=\text{exp} (2\pi i/n)$)
If $n$ is odd then $\gcd (2,n)=1$. Hence, \begin{align*} \prod_{j=1}^{\frac{n-1}{2}} (\zeta_n^{t j}+\zeta_n^{-t j}+2)&=\prod_{j=1}^{\frac{n-1}{2}} (\zeta_n^{2t j}+\zeta_n^{-2t j}+2)\\ &=\prod_{j=1}^{\frac{n-1}{2}}(\zeta_n^{t j} +\zeta_n^{-t j})^2\\ &=\Bigg[ \prod_{j=1}^{\frac{n-1}{2}} (\zeta_n^{t j}+\zeta_n^{-t j}) \Bigg]^2\\ &=\prod_{j=1}^{n-1}(\zeta_n^{t j}+\zeta_n^{-t j})\\ &= (\underbrace{\zeta_n \zeta_n^2 ... \zeta_n^{n-1}}_{=1})^{t} \prod_{j=1}^{n-1}(\zeta_n^{t j}+\zeta_n^{-t j})\\ &=\prod_{j=1}^{n-1} \zeta_n^{t j}(\zeta_n^{t j}+\zeta_n^{-t j})\\ &=\prod_{j=1}^{n-1} (1+\zeta_n^{2t j}) \\ &=\prod_{j=1}^{n-1} (1+\zeta_n^{t j}). \end{align*}
But \begin{align*} \zeta_n^{t j}=1 &\iff n\mid t j \\ &\iff \frac{n}{(t,n)} \mid \frac{t}{(t,n)}j \\ &\iff \frac{n}{(t,n)} \mid j \\ &\iff j\in \bigg\{ \frac{n}{(t,n)},\frac{2n}{(t,n)},...,\frac{((t,n)-1)n}{(t,n)} \bigg\}. \end{align*}
So, \begin{align*} \prod_{j=1}^{n-1} (1+\zeta_n^{t j}) = 2^{(t,n)-1}\Bigg( \prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j} \neq 1}} (1+\zeta_n^{t j}) \Bigg). \end{align*}
Moreover, \begin{align*} \Bigg( \prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j}\neq 1}} (1-\zeta_n^{t j}) \Bigg)\Bigg( \prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j}\neq 1}} (1+\zeta_n^{t j}) \Bigg) = \prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j}\neq 1}} (1-\zeta_n^{2 t j}) =\prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j}\neq 1}} (1-\zeta_n^{t j}), \end{align*} since $\{ \zeta_n^{2t j} \colon \zeta_n^{t j}\neq 1, \, 1\leq j\leq n-1 \}=\{ \zeta_n^{t j} \colon \zeta_n^{t j}\neq 1 , \, 1\leq j\leq n-1\}$. Indeed, let $\zeta_n^{2t j}$ be an arbitrary element of the first set. Then, $n\nmid 2j$ since otherwise we would have $n \mid j$ and therefore $\zeta_n^{t j}=1$ (which is not true). Thus, let $l\in \{1,...,n-1\}$ such that $2j \equiv l \, (\text{mod} \, n)$. Then $\zeta_n^{2tj}=\zeta_n^{t (2j)}=\zeta_n^{t l}$. For the other inclusion, just notice that $\zeta_n^{tj}=\zeta_n^{2t l}$, where $l=\frac{j}{2}$ if $j$ is even and $l=\frac{n+j}{2}$ otherwise.
Now, \begin{equation*} \prod_{\substack{1\leq j\leq n-1 \\ \zeta_n^{t j}\neq 1}} (1+\zeta_n^{t j})=1, \end{equation*} which concludes the demonstration.