<s>[INST] Proving that sequence $a_n = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}} = x^{1-2^{-n}}$ Let $x>0$.
For sequence $a_n$, such that $n$ denotes the $n$th term:
$$\begin{align} a_1&=  \sqrt{x}\\
a_2&= \sqrt{x \sqrt{x}}\\
a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\
a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\
&\vdots\\
a_{n-1}&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x}}}}\\
a_n&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x \sqrt{x}}}}}\end{align}$$
How could one prove that:
$${a_n =  x^{1-2^{-n}}}?$$
 [/INST] Hint: 
\begin{align}
a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}}
\end{align}
Can you use induction in these footsteps?
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