Prove or disprove : $\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256...}}}} = \sqrt{2+\sqrt{5}}$ Edit: My initial question was regarding the expressions:
$\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{64... + \sqrt{4^{n}}}}}}$
And in general:
$\sqrt{1 + \sqrt{k+\sqrt{k^2+\sqrt{k^3... + \sqrt{k^{n}}}}}}$
And their limits, however, my working out was wrong. So, if someone can explain how to find the values of the above two expressions, that would be great!
This is just something I've noticed playing around with the radicals, and I honestly don't have much idea on how to prove it. These were my ideas:
Let $\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}} = A_n$
My logic was as follows:
$A_n = \sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}}$
$A_n^2 = 1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\frac{1}{4}\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\sqrt{1+\frac{1}{16}\sqrt{256+... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\sqrt{1+\sqrt{1+... + \sqrt{1}}}}$
$A_{\infty}^2 = 1 + 2\sqrt{1+\sqrt{1+\sqrt{1+... }}} = 1 + 2\phi = 2 + \sqrt{5}$
$A_{\infty} = \sqrt{2 + \sqrt{5}}$
Then, rather than this specific case, how would we evaluate:
$L = \sqrt{1 + \sqrt{k+\sqrt{k^2+\sqrt{k^4+\sqrt{k^8+...}}}}}$
Would we also get $L = \sqrt{\sqrt{k}\phi + 1}$?
And what about expressions such as:
$\sqrt{1 + \sqrt{2+\sqrt{3+...}}}$
Thanks for any answers and guidance!

Let $\phi_n:=\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}$ ($n$ square roots); note that $\phi_n\to\phi$, the golden ratio.
Let $a_{n,r}:=\sqrt{k^{2^{r-n+1}}+\sqrt{\cdots+\sqrt{k^{2^r}}}}$.
For induction in $n$, assume that $a_{n,r}=k^{2^{r-n}}\phi_n$.
Then $$a_{n+1,r}=\sqrt{k^{2^{r-n}}+a_{n,r}}=\sqrt{k^{2^{r-n}}+k^{2^{r-n}}\phi_n}=k^{2^{r-n-1}}\phi_{n+1}$$
The initial case $n=0$ is trivially true.
Thus $\sqrt{k+\sqrt{\cdots}}=\lim_{r\to\infty}a_{r+1,r}=\lim_{r\to\infty}\sqrt{k}\phi_{r+1}=\sqrt{k}\phi$.
Hence $\sqrt{1+\sqrt{k+\cdots}}=\sqrt{1+\sqrt{k}\phi}$.
