Write $a = \alpha^{-2}$. and $p = 2017$, which is an (odd!) prime. Then your product can be written as
$$ \alpha ^{p\cdot (p-1)/2 } N,\text{ where } N= \prod_{r=1}^{p-1} (1 + a^r).$$
The first term equals $+1$, since $p(p-1)/2= 2017\cdot 1008$ is even.
For the second term... Set
$$\phi(x) ={x^p-1 \over x-1}= x^{p-1}+\cdots + 1.$$
(Consider the first equality as 'definition', and the second as arising from long division. FWIW, $\phi$ is called the $p$th 'cyclotomic' polynomial.)
Evaluating $\phi(x)= x^{p-1}+\cdots + 1$ at $x=1$, one gets
$\phi(1) = p$.
On the other hand, If $b^p=1$, but $b^r\not =1 $, for any $1\le r<p$ ( e.g., $b=e^{2\pi i /p}$ - any such $b$ is called a 'primitive' $p$th root of unity),
all of the zeros/roots of $x^p-1$ are of the form $b^r$, where $r=0,\cdots, p-1$, therefore
$$ \phi(x) = {\prod_{r=0}^{p-1} (x- b^r) \over (x-1)} = \prod_{r=1}^{p-1} (x- b^r).$$
Therefore, (from the earlier evaluation),
$$\phi(1) = \prod_{r=1}^{p-1} (1- b^r) = p.$$
So, using the fact that $a$ (defined in the first line of this post) is a $p$th primitive root of unity,
one has that
$$ P = \prod_{r=1}^{p-1} ( 1-a^r) = p.$$
Likewise, $$NP = \prod_{r=1}^{p-1} (1- (a^2)^r) =p,$$
as $a^2$ is also a primitive pth root of unity. Hence, (earning me a million dollars of Clay prize money and possibly even a job somewhere), $NP=p$. Canceling the $p$'s, one gets $N =1$.
Therefore, the original product is also equal to one.
BTW, the above evaluating of $N$ only really seems to use only that $p$ is odd; one does not need that $p$ be a prime. On the other hand, $\phi(x)$ would not be 'the' cyclotomic polynomial.
