jmc

Notice that $$\frac{2^{2^k}}{4^{2^k}-1}=\frac{2^{2^k}+1}{4^{2^k}-1}-\frac{1}{4^{2^k}-1}=\frac{1}{2^{2^k}-1}-\frac{1}{4^{2^k}-1}=\frac{1}{4^{2^{k-1}}-1}-\frac{1}{4^{2^k}-1}.$$Therefore, the sum telescopes as $$\left(\frac{1}{4^{2^{-1}}-1}-\frac{1}{4^{2^0}-1}\right)+\left(\frac{1}{4^{2^0}-1}-\frac{1}{4^{2^1}-1}\right)+\left(\frac{1}{4^{2^1}-1}-\frac{1}{4^{2^2}-1}\right)+\cdots$$and evaluates to $1/(4^{2^{-1}}-1)=\boxed{1}$.