Why do we have
$$ \ln(z) = \frac{z-1}{e -1} \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {z^{2^{-n}} + 1} $$
Is it possible to show that the derivative of the product is $z^{-1}$ without showing that the product is the logarithm ?
Does that imply there exists a solution set $a_n$ such that :
$$ \ln(z) = \sum (-1)^n a_n z^{1/n} $$
??