Let P_n = _ k=0 ^ n-1 ( (2^k π)/(129) ) . If P_n = (1)/(2^n) , then...

Given $P_n = \prod_{k=0}^{n-1} \cos \left( \dfrac{2^k \pi}{129} \right)$

Using the standard identity for the product of cosines with angles in a geometric progression:
$\prod_{k=0}^{n-1} \cos(2^k \theta) = \dfrac{\sin(2^n \theta)}{2^n \sin \theta}$

Substitute $\theta = \dfrac{\pi}{129}$:
$P_n = \dfrac{\sin \left( \dfrac{2^n \pi}{129} \right)}{2^n \sin \left( \dfrac{\pi}{129} \right)}$

We are given that $P_n = \dfrac{1}{2^n}$. Equating the two expressions:
$\dfrac{\sin \left( \dfrac{2^n \pi}{129} \right)}{2^n \sin \left( \dfrac{\pi}{129} \right)} = \dfrac{1}{2^n}$

$\Rightarrow \sin \left( \dfrac{2^n \pi}{129} \right) = \sin \left( \dfrac{\pi}{129} \right)$

The general solution for $\sin x = \sin y$ is $x = m\pi + (-1)^m y$, where $m$ is an integer.
$\dfrac{2^n \pi}{129} = m\pi + (-1)^m \dfrac{\pi}{129}$

Dividing by $\pi$ and multiplying by $129$:
$2^n = 129m + (-1)^m$

Case 1: $m$ is an even integer, say $m = 2k$.
$2^n = 129(2k) + 1 = 258k + 1$
Since $2^n$ is even for $n \ge 1$ and $258k + 1$ is always odd, this case yields no positive integer solutions for $n$.

Case 2: $m$ is an odd integer, say $m = 2k + 1$.
$2^n = 129(2k + 1) - 1 = 258k + 129 - 1 = 258k + 128$

To find the smallest positive integer $n$, we test non-negative integer values for $k$:
For $k = 0$:
$2^n = 128$
$\Rightarrow 2^n = 2^7 \Rightarrow n = 7$

For $k = 1$:
$2^n = 258(1) + 128 = 386$ (which is not a power of $2$).

Thus, the smallest positive integer $n$ satisfying the equation is $7$.

Answer: $7$