Is That A Long Division Sign?

Relevant wiki: Exponential Inequalities

Squaring all of the numbers twice, we are finding $4,4,\sqrt{8},2$ . Thus the last one, $\sqrt{\sqrt{\sqrt{\sqrt{2\times2\times2\times2}}}}$ , is the smallest.

Relevant wiki: Simplifying Expressions with Radicals - Basic

$\sqrt2=2^{1/2}$ .

$\sqrt{\sqrt{2\times2}}=2^{1/2}$ .

$\sqrt{\sqrt{\sqrt{2\times2\times2}}}=2^{3/8}$ .

$\sqrt{\sqrt{\sqrt{\sqrt{2\times2\times2\times2}}}}=2^{1/4}$ .

Bases are same in all the cases. Comparing the powers we are that $\dfrac 1 4$ is the smallest so

$\sqrt{\sqrt{\sqrt{\sqrt{2\times2\times2\times2}}}}$ is the smallest.

$\sqrt{2}=2^{\dfrac{1}{2}}=2^{\dfrac{4}{8}}$

$\sqrt{\sqrt{2×2}}=4^{\dfrac{1}{4}}=2^{\dfrac{4}{8}}$

$\sqrt{\sqrt{\sqrt{2×2×2}}}=2^{\dfrac{3}{8}}$

$\sqrt{\sqrt{\sqrt{\sqrt{2×2×2×2}}}}=2^{\dfrac{2}{8}}$

Now, we can know by just seeing that which one is the smallest, and that is $\boxed{\sqrt{\sqrt{\sqrt{\sqrt{2×2×2×2}}}}}$