Expanding Seems Implausible

For $j=1$ note that $\displaystyle \prod_{i=1}^{j} {\omega_j}^{2i-1} = ({\omega_1 - 1} )= -2$ as $\omega_1 = -1$ .

For $j=2$ expansion of the product gives $\displaystyle \prod_{i=1}^{j} {\omega_j}^{2i-1} = (i-1)(-i-1) = 2$ .

Similarly for $j=3$ we have $\displaystyle \prod_{i=1}^{j} {\omega_j}^{2i-1} = (\omega_3-1)(\omega_3^{3} - 1)(\omega_3^{5} - 1) = (-2)(\omega_3-1)(\omega_3^{5} - 1) = (-2) (\omega_3-1)(-\omega_3^{2} - 1) \\ = (-2) (\omega_3^{6} - \omega_3 + \omega_3^{2} + 1) = -2 .$

The last equality follows from the relation ( $1 - \omega_3 + \omega_3^{2}) = 0$ because $\omega_3^{3} = -1$ .

Following similar argument we see that $\displaystyle |\prod_{j=1}^{2014} \prod_{i=1}^{j} ({\omega_j}^{2i-1}-1)| = |(-2) \times 2 \times (-2) \times \dots \times 2| = 2^{2014}$ .

For any $j$ , define a polynomial $P(x)$ with roots $\omega_j^{2i-1}-1$ for $1 \leq i \leq j$ . We have that ${(\omega_j^{2i-1})}^j=(\omega_j^j)^{2i-1}=(-1)^{2i-1}=-1$ . Therefore, ${(\omega_j^{2i-1}-1+1})^j+1=0$

Suppose $r^j=-1$ for some $r$ . Then $r=(e^{i\pi/j})^{2n-1}$ for some $n$ such that $1 \leq n \leq j$ . Therefore, $P(x)=(x+1)^j+1$ because the polynomial is 0 at these points and no other points (If there was another root, setting $r=x+1$ would quickly give a contradiction).

The product of the roots of a polynomial is the constant term if the degree of the polynomial is even and the negative of the constant term if the degree is odd. The constant term of this polynomial for any $j$ is clearly 2. Therefore, the product for and $j$ is 2 if $j$ is even and -2 if $j$ is odd.

We are now looking for the product $\prod _{ j=1 }^{ 2014 }{ { (-2) }^{ j } }$ , which is clearly the same as $2^{2014}$ . We now have $a=2$ and $b=2014$ , so $a+b=\boxed{2016}$ .