What Happens As We Go Deeper?

Considering $\sqrt{x}-\sqrt{\sqrt{x}}=x^{\frac{1}{2}}-x^{\frac{1}{4}}=x^{\frac{1}{4}}(x^2-1)<0 \ \ (Since \ x<0)$

With $\sqrt{\sqrt{x}}-\sqrt{\sqrt{\sqrt{x}}}$ and $\sqrt{\sqrt{\sqrt{x}}}-\sqrt{\sqrt{\sqrt{\sqrt{x}}}}$ we get the same results $\therefore \sqrt{\sqrt{\sqrt{\sqrt{x}}}}>\sqrt{\sqrt{\sqrt{x}}}>\sqrt{\sqrt{x}}>\sqrt{x}$

We start by proving that, if $0<x<1$ , then $x<\sqrt{x}$ :

Notice that for any number $x$ , such that $0<x<1$ , $x^2<x$ (a fraction of a positive real number is evidently smaller than the positive number itself). If we now take the square root on both sides, we obtain: $x<\sqrt{x}$

If we again take the square root on both sides 3 times, we obtain 3 more inequalities:

$\sqrt{x}<\sqrt{\sqrt{x}}$

$\sqrt{\sqrt{x}}<\sqrt{\sqrt{\sqrt{x}}}$

$\sqrt{\sqrt{\sqrt{x}}}<\sqrt{\sqrt{\sqrt{\sqrt{x}}}}$

Now we combine all 4 inequalities and it follows that: $\sqrt{x}<\sqrt{\sqrt{x}}<\sqrt{\sqrt{\sqrt{x}}}<\sqrt{\sqrt{\sqrt{\sqrt{x}}}}$

Q.E.D