We Need To Go Deeper - 2

Theorem:

If $x>1$ and $a>b$ , we have

$x^a>x^b$

Example: $x=2,\;a=4,\;b=2\implies x^a=2^4=16,\;x^b=2^2=4,\implies16>4$

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Now, we convert all the roots into indices:

$\sqrt{x}=x^{1/2}\\ \sqrt{\sqrt{x}}=\left(x^{1/2}\right)^{1/2}=x^{1/2 \times 1/2}=x^{1/4}\\ \sqrt{\sqrt{\sqrt{x}}}=\left(\left(x^{1/2}\right)^{1/2}\right)^{1/2}=x^{1/2 \times 1/2 \times 1/2}=x^{1/8}\\ \sqrt{\sqrt{\sqrt{\sqrt{x}}}}=\left(\left(\left(x^{1/2}\right)^{1/2}\right)^{1/2}\right)^{1/2}=x^{1/2 \times 1/2 \times 1/2 \times 1/2}=x^{1/16}$

Since $x>1$ and $\dfrac{1}{2}>\dfrac{1}{4}>\dfrac{1}{8}>\dfrac{1}{16}$ , we have

$x^{1/2}>x^{1/4}>x^{1/8}>x^{1/16}\\ \sqrt{x}>\sqrt{\sqrt{x}}>\sqrt{\sqrt{\sqrt{x}}}>\sqrt{\sqrt{\sqrt{\sqrt{x}}}}$

Therefore, the smallest number is $\boxed{\sqrt{\sqrt{\sqrt{\sqrt{x}}}}}$