Radical Radicals

It's easy to see that the elements of $S_1$ are the solutions to $x^2 - 2 = 0$ . Likewise, the elements of $S_2$ are the solutions of $(x^2 - 2)^2 -2 = 0$ . A simple induction shows that the elements of $S_n$ are the solutions to $(((\cdots (x^2 -2)^2 - 2)^2 - 2)^2 - \cdots 2)^2 - 2 = 0$ , where the squares (powers of 2) were used exactly $n$ times.

Let $f_n (x) = \underbrace{(((\cdots (x^2 -2)^2 - 2)^2 - 2)^2 - \cdots 2)^2}_{n \text{ squares}} - 2$ .

If $a_1, a_2, \ldots$ are all the roots of $f_n (x) = 0$ , then $-(a_1 + 1), -(a_2 + 1), -(a_3 + 1) , \ldots$ are all the roots of $f_n (-(x-1)) = 0$ .

So, $f_n (-(x-1)) = \underbrace{(((\cdots ((x+1)^2 -2)^2 - 2)^2 - 2)^2 - \cdots 2)^2}_{n \text{ squares}} - 2$ .

By Vieta's formula, $\displaystyle \prod_{a\in S_n} (1-a)$ is simply the coefficient of the constant of the polynomial $f_n (-(x-1))$ .

This value, call it $c_{n}$ , satisfy the recursive relation $c_1 = -1, c_{n+1} = (c_n)^2 - 2$ . A(nother) induction shows that $c_1 = c_2 = c_3 = c_4 = \cdots = -1$ .

Hence, our answer is just $c_{2017} = \boxed{-1}$ .