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More generally, consider $$\prod _{k=0}^{\infty }(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)\cdots .$$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, $$(1+x)(1+x^2)(1+x^4)\cdots =\frac{1-x^2}{1-x}\cdot \frac{1-x^4}{1-x^2}\cdot \frac{1-x^8}{1-x^4}\cdots =\frac{1}{1-x}.$$For $x=\frac{1}{14},$ this is $\frac{1}{1-\frac{1}{14}}=\boxed{\frac{14}{13}}.$