If $$f(x) = \sum_{r=0}^{20}\binom{20}{r}\big(\ln(1+x)\big)^r$$ Find $$\sum_{r=0}^\infty \bigg(h(x^{2^r})-1\bigg)$$ where, $h(x) = \big(f(x)\big)^{1/20}$
My Attempt:
$f(x)$ is the binomial expansion of $(1+\ln(1+x))^{20}$.
Then, $h(x) = 1+\ln(1+x)$
Plugging this into the summation, I get
$$\sum_{r=0}^\infty \ln(1+x^{2^r})$$
I am unable to go any further. How do you find this?
Any help would be appreciated.
Hint
For
$1-x\ne0,$
$$(1-x)\prod_{r=0}^n(1+x^{2^r})=1-x^{2^{n+1}}$$
As $\ln(1+x)$ is defined in $(-1,+1]$
we need to consider $x=1$ separately
Otherwise for $|x|<1,$ $$\lim_{n\to\infty}x^{2^{n+1}}=?$$