If $x\geq 0$, then $\sqrt{x\sqrt{x\sqrt{x}}}=$
$\textbf{(A) } x\sqrt{x}\qquad \textbf{(B) } x\sqrt[4]{x}\qquad \textbf{(C) } \sqrt[8]{x}\qquad \textbf{(D) } \sqrt[8]{x^3}\qquad \textbf{(E) } \sqrt[8]{x^7}$
Recall that square roots are one-half powers, namely $\sqrt y=y^{\frac12}$ for all $y\geq0.$
We have
\begin{align*} \sqrt{x\sqrt{x\sqrt{x}}} &= \sqrt{x\sqrt{x\cdot x^{\frac12}}} \\ &= \sqrt{x\sqrt{x^{\frac32}}} \\ &= \sqrt{x\cdot\left(x^{\frac32}\right)^{\frac12}} \\ &= \sqrt{x^{\frac74}} \\ &= \left(x^{\frac74}\right)^{\frac12} \\ &= x^{\frac78} \\ &= \boxed{\textbf{(E) } \sqrt[8]{x^7}}. \end{align*}
~Hapaxoromenon (Solution)
~MRENTHUSIASM (Reformatting)