Contradicting Nested Radicals I encountered a question that puzzles me. The task is to find the simplified form of $$x\sqrt[2]{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7] {x}}}}}}$$ my answer is $$x^\frac{433}{252}$$ but as I graph the unsimplified nested radical to desmos grapher, the graph covers all real numbers as its domain and all real numbers as its range. But as I graph $$x^\frac{433}{252}$$ the graph only assumes all positive numbers as its domain and $$[0,+infinity)$$ as its range... I check the given and I notice something. If I give a negative value for x,, then evaluates the unsimplified nested radical from the inside, the sign follows the parity of the index... it means that it was smoothly evaluated through all indexes.. resulting to a negative final result. But if I give the same negative value of x to the simplified form that is $$x^\frac{433}{252}$$ the even power will make the given number positive thus the final result will be positive. It contradics theprevious result. I believe I'm missing a fundamental concept here. Any idea would be a great help. Tnx!!
 A: Suppose $x<0$; then


*

*$\sqrt[7]{x}<0$

*$x\sqrt[7]{x}>0$

*$\sqrt[6]{x\sqrt[7]{x}}>0$

*$x\sqrt[6]{x\sqrt[7]{x}}<0$

*$\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}<0$

*$x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}>0$

*$\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}>0$

*$x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}<0$

*$\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}}<0$

*$x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}}>0$

*$\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}}}>0$

*$x\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x\sqrt[7]{x}}}}}}<0$


So you can define the expression for any $x$, and the result is the same as
$$
|x|^{433/252}\operatorname{sgn}x
$$
A: The identity $\sqrt[n]{x} = x^{1/n}$ is only valid for nonnegative $x$, exactly because otherwise you run into problems like the one you have here.
In fact, even definining $x^a$ with $a\notin \Bbb Z$ is tricky for negative bases, so comparing it to some root of $x$, which may or may not exist, remains a dubious prospect.
