Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$

I'd rather try this:
If
$$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$
then 
$$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\
\end{align}$$
Thus if $u=\frac1{\sqrt 2}$ then 
$$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$
The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.
