How to do analytic continuation for following function?
$$f(z) = \prod\_{n=0}^{+\infty} {(1+z^{4^n})}$$
Evidently it satisfies $f(z)f(z^2)=\dfrac{1}{1-z}$...



Analytic continuation for PI(1+z^(4^n))
All the answers above are absolutely fine. Just to give a completely obvious argument: 


Take a the line $z=t\exp(i\phi)$ with $\phi=2\pi/{4^{n\_0}}$ for some $n\_0\in \mathbb{N}$.
Then 
$$
f(z)=\Pi\_{n=0}^{n\_0-1}\left(1+t^{4^n}\exp\left(i\phi^{4^n}\right)\right)\Pi\_{n=n\_0}^{\infty}(1+t^{4^n}) 
$$
Since $4^{n\_0}/2\neq 4^{n^\prime}$, $\Pi\_{n=0}^{n\_0-1}\left(1+\exp\left(i\phi^{4^n}\right)\right)\neq0$, and therefore $f$ diverges at $t=1$. 


So your analytic extension would have an accumulation of essential singular points, on the sequence
$$
\left\{\exp\left(i\frac{2\pi}{4^n}\right),1\right\},
$$ 
and that's not very popular in the analytic community.


