The answer is 1. To see that, notice first that as r goes from 0 to 5, the set of values of is the same as the set of values of as k goes from 1 to 6. In fact, modulo 13 we have , , , , , . So for r=0,1,2,4,5, for k = 1,2,4,3,6 respectively; and using the fact that , we see that the remaining value is equal to .

Next, notice that is positive for k=1,2,3,4 and negative for k=5,6. So the product is positive. Also, (because of that same fact as in the previous paragraph). So the prduct we are looking for is the positive square root of .

Let . Then . The numbers , together with 1, are the roots of the equation , so their product is 1. Also, . But as k goes from 1 to 12, the complex numbers run through the same set of values as the numbers . So the terms in the numerator all cancel with terms in the denominator, and we are left with