Find
\[\prod_{k = 0}^\infty \left( 1 + \frac{1}{14^{2^k}} \right).\]
More generally, consider
\[\prod_{k = 0}^\infty (1 + x^{2^k}) = (1 + x)(1 + x^2)(1 + x^4) \dotsm.\]where $x < 1.$  (The product in the problem is the case where $x = \frac{1}{14}$.)

We can write
\[1 + x^{2^k} = \frac{(1 + x^{2^k})(1 - x^{2^k})}{1 - x^{2^k}} = \frac{1 - x^{2^{k + 1}}}{1 - x^{2^k}}.\]Hence,
\[(1 + x)(1 + x^2)(1 + x^4) \dotsm = \frac{1 - x^2}{1 - x} \cdot \frac{1 - x^4}{1 - x^2} \cdot \frac{1 - x^8}{1 - x^4} \dotsm = \frac{1}{1 - x}.\]For $x = \frac{1}{14},$ this is $\frac{1}{1 - \frac{1}{14}} = \frac{14}{13}.$