This post is inspired by this question Justifying $\sum_{n=0}^\infty\log(1+x^{2^n}) = -\log(1-x)$ for $0\le x<1$
Also this is the crucial step in the second official solution (justified differently there) to Putnam 2016 problem B6.
A Power-Series Proof
We know that $$\ln(1+t)=\sum_{m=1}^{\infty}(-1)^{m-1}\frac{t^m}{m}$$ Therefore $$\ln(1+x^{2^n})=\sum^{\infty}_{m=1}(-1)^{m-1}\frac{x^{2^nm}}{m}$$ Hence $$\sum_{n=0}^{\infty}\ln(1+x^{2^n})=\color{red}{\sum_{n=0}^{\infty}\sum^{\infty}_{m=1}(-1)^{m-1}\frac{x^{2^nm}}{m}=\sum^{\infty}_{m=1}\sum_{n=0}^{\infty}(-1)^{m-1}\frac{x^{2^nm}}{m}}$$ Consider the series $$f(z):=\sum_{n=0}^{\infty}z^{2^n}$$ Then we get $$\sum_{n=0}^{\infty}\ln(1+x^{2^n})=\sum^{\infty}_{m=1}\frac{(-1)^{m-1}}{m}f(x^m)$$
Note that $$f(z^2)=f(z)-z\tag{1}$$ Denote $$E=\sum_{k=1}^{\infty}\frac{f(x^{2k})}{2k}$$ and $$O=\sum_{k=1}^{\infty}\frac{f(x^{2k-1})}{2k-1}$$ Then $$\sum_{n=0}^{\infty}\ln(1+x^{2^n})=O-E\tag{2}$$ Again from $(1)$ we get
$$E=\sum_{k=1}^{\infty}\frac{f(x^{2k})}{2k}=\sum_{k=1}^{\infty}\frac{f(x^k)-x^k}{2k}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{f(x^k)}{k}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{x^k}{k}=\frac{O+E}{2}+\frac{\ln(1-x)}{2}$$ Therefore $$\sum_{n=0}^{\infty}\ln(1+x^{2^n})=O-\frac{O+E}{2}-\frac{\ln(1-x)}{2}=\frac{O-E}{2}-\frac{\ln(1-x)}{2}$$ $$\implies\sum_{n=0}^{\infty}\ln(1+x^{2^n})=\frac{\sum_{n=0}^{\infty}\ln(1+x^{2^n})}{2}-\ln(1-x)\tag{from (2)}$$ $$\implies\sum_{n=0}^{\infty}\ln(1+x^{2^n})=-\ln(1-x)\tag*{$\square$}$$
Edit
Since, for $x\in[0,1)$, the double series $$\sum_{n=0}^{\infty}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{x^{2^nm}}{m}$$ is absolutely convergent, we can split the sum in the following way and can observe why is it possible to swap the order of summation in the $\color{red}{\text{red}}$ part above
$$\sum_{n=0}^{\infty}\sum^{\infty}_{m=1}(-1)^{m-1}\frac{x^{2^nm}}{m}=\sum_{n=0}^{\infty}\left(-\sum^{\infty}_{k=1}\frac{x^{2^n2k}}{2k}+\sum^{\infty}_{k=1}\frac{x^{2^n(2k-1)}}{(2k-1)}\right)\\=-\sum_{n=0}^{\infty}\sum^{\infty}_{k=1}\frac{x^{2^n2k}}{2k}+\sum_{n=0}^{\infty}\sum^{\infty}_{k=1}\frac{x^{2^n(2k-1)}}{(2k-1)}\\=-\sum^{\infty}_{k=1}\sum_{n=0}^{\infty}\frac{x^{2^n2k}}{2k}+\sum^{\infty}_{k=1}\sum_{n=0}^{\infty}\frac{x^{2^n(2k-1)}}{(2k-1)}$$ We can interchange the order of summation due to the non-negativity of the summands and hence it becomes, $$=\sum^{\infty}_{m=1}\sum_{n=0}^{\infty}(-1)^{m-1}\frac{x^{2^nm}}{m}$$
We have used Tonelli's theorem for changing the order of infinite sums with non-negative summands.