How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$ How to solve this equation?
$$x=\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}.$$
Answer: $x=2$
 A: From this paper, on page $5$, equation $(20)$ and letting $x=1$ one sees that the value of your nested radical is indeed $2$.
This comes from the following:
Let $a:=\{a_n\}_{n\geq 0}$ and $b:=\{b_n\}_{n\geq 0}$ be sequences. We will make use of the following notation:
$$R(a,b):=\sqrt{a_0 + b_0\sqrt{a_1 + b_1\sqrt{a_2 + b_2\sqrt{\ldots}}}}$$

Theorem:
Let $x_1,x_2,x_3 \geq 0$
We define the sequences $p,q,r,s$ where:
$$p_n = x_1x_2 + (x_2 + 2^nx_3)^2 \qquad q_n = x_1 $$
$$r_n = x_2(x_1 + nx_3) + (x_2 + x_3)^2\qquad s_n = x_1 + nx_3$$
Then
$$x_1 + x_2 + x_3 = R(p,q) = R(r,s).$$

The proof is in the article and it's very interesting, but I shall not transcribe it here.
If you consider the sequences $p$ and $q$ for $x_1 = x_3 = 1$ and $x_2 = 0$ you get the desired result.
A: Hint: Can you find a relation between $x$ and $x^2$?
A: just an attempt to reduce the size of the numbers...
$$
\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}=\sqrt {1+ 2^1\sqrt{1+2^0\sqrt {1+2^{-1}\sqrt {1+2^{-2}\sqrt {1+2^{-3}\sqrt{1+\ldots }}}}}}
$$
A: I found this answer.Is that wrong ? 

