$u_i$ are complex roots of 1, but of n+1 degree (excluding $u_0$ = 1)
My attempts, I don't know if it leads anywhere or what to do next: We can tell that $u_i = (u_1)^{i}$ so we have: $(2 - u_1)(2 - u_1^2)...(2 - u_1^2)$ Supposing n+1 is odd number (even case is similar), we can tell that half of this roots is equal to its conjugate, so: $ 2 - u_i = \overline{2 - u_{n-i}}$. Because $z\overline{z} = |z|$ are equation reduces to: $|2 - u_1|^2 |2 - u_1^2|^2 \dots |2 - u_1^{\frac{n+1}{2}}|$
So this means imaginary part is 0? What about real part? I don't know what I can do next or this is all wrong.
You have that the $u_i$'s are the roots of $$ \frac{x^{n+1}-1}{x-1}=x^n+...+x+1=(x-u_1)\cdot ... \cdot (x-u_n)=p(x) $$ Therefore you want to calculate $p(2)$ that is equal to $$ \sum_{i=0}^n2^i=\frac{2^{n+1}-1}{2-1}=2^{n+1}-1 $$