Given that $w_1, w_2, w_3, \ldots , w_{13}$ are distinct $13^{th}$ roots of unity .

Find the value of $\displaystyle \prod_{k=1}^{13} (2-w_k)$ .

The answer is 8191.

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For the 13th root of unity $\omega$ , we have:

$\begin{aligned} (x - \omega_1)(x - \omega_2)(x - \omega_3) \cdots (x - \omega_{13}) & = x^{12} + x^{11} + x^{10} + \cdots + x + 1 \\ \implies \prod_{k=1}^{13} (2-\omega_k) & = \sum_{k=0}^{12} 2^k = \frac {2^{13}-1}{2-1} = \boxed{8191} \end{aligned}$