<s>[INST] How can we simplify the expression $P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$? Is there a way to reduce the expression
$P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$?
 [/INST] So our target expression is:
$$ P + \sqrt{P^2 + \sqrt{P^4...}}$$
An indirect approach that is quite simple is to consider
$$ P \left( 1 + \sqrt{1 + \sqrt{1 ...}} \right)$$
It follows that of we distribute the P we will have 
$$ P +P \sqrt{1 + \sqrt{1 ...}} $$
Which yields
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
And again we can distribute under the next root for $P^4$ then $P^8$ etc...
So now the trick is to evaluate:
$$S = 1+\sqrt{1+\sqrt{...}}$$
It follows that 
$$(S-1)^2 = 1+\sqrt{1+\sqrt{...}}$$
And therefore 
$$ (S-1)^2 = S$$
Using this we can solve and find
$$S^2-3S + 1 = 0$$
Which has solutions given by the quadratic formula:
$$S = \frac{3 \pm \sqrt{5}}{2}$$
A simple test of evaluating the first few terms of 
$$S = 1+\sqrt{1+\sqrt{...}}$$
Suggests that solution closer to $2$ is in order this we have 
$$S =  \frac{3 + \sqrt{5}}{2}$$
And this to evaluate:
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
Is just
$$P \left(   \frac{3 + \sqrt{5}}{2}   \right)$$
</s>