<s>[INST] Nested Square Roots $5^0+\sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt\dots}}}$ How would one go about computing the value of $X$, where
$X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$
I have tried the standard way of squaring then trying some trick, but nothing is working. I have also looked at some nested radical previous results, but none seem to be of the variety of this problem. Can anyone come up with the answer? Thank 
 [/INST] The trick is to pull out a $\sqrt{5}$ factor from the second term:
$$
\frac{\sqrt{5^1+ \sqrt{5^2 + \sqrt{5^4 + \sqrt{5^8 + \cdots}}}}}{\sqrt{5}} 
= \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}, 
$$
which I call $Y$ for convenience. To see why this is true, observe that 
$$ \begin{align*} \frac{\sqrt{5^1+x}}{\sqrt{5}} = \sqrt{1+\frac{x}{5^1}} \\ 
 \frac{\sqrt{5^2+x}}{5^1} = \sqrt{1+\frac{x}{5^2}} \\
 \frac{\sqrt{5^4+x}}{5^2} = \sqrt{1+\frac{x}{5^4}} \end{align*}
$$
and so on. Applying these repeatedly inside the nested radicals gives the claim. (The wording of this explanation has been inspired mostly by a comment of @J.M. below.) 
Now, it only remains to compute $Y$ and $X$. By squaring, we get 
$$
Y^2 = 1 + Y \ \ \ \ \implies  Y = \frac{\sqrt{5}+1}{2},
$$
discarding the negative root. Plugging this value in the definition of $X$, we get:
$$
X = 1 + \sqrt{5} Y = 1 + \frac{5+\sqrt{5}}{2} = \frac{7+\sqrt{5}}{2} .
$$
Note on the convergence issues. As @GEdgar points out, to complete the proof, I also need to demonstrate that both sides of the first equation do converge to some finite limits. For our expressions, convergence follows from @Bill Dubuque's answer to my question on the defining the convergence of such an expression. I believe that with some work, one can also give a direct proof by showing that this sequence is bounded from above (which I hope will also end up showing the theorem Bill quotes), but I will not pursue this further. Added: See @Aryabhata's answer to a related question for a hands-on proof.
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