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
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$

Just a comment on such similar cases. From this short answer,
$$x+2^n=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\tag{1}$$
Let $N=\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$. Then
$$\begin{aligned}
 N &< \sqrt{1+\sqrt{2+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}} \\
 N &< \sqrt{1+\sqrt{5}} \\
  &< 1.79890743995\\
\end{aligned}$$
For comparison, $N ≃ 1.78316580926410..$. Also from Eq. $1$, we can get;
$$\boxed{1+\sqrt{2}=\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}}$$
$$\begin{aligned}
N &< \sqrt{1+\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}} \\
 &< \sqrt{2+\sqrt{2}} \\
\end{aligned}$$
