Four Fours/Lemmata/Two Fours/64/Solutions/2

Solution

 * $64 = \sqrt {\paren {\sqrt {\sqrt 4} }^{4!} }$

Proof

 * $\sqrt {\paren {\sqrt {\sqrt 4} }^{4!} } = \sqrt {\sqrt {2^{24} } } = \sqrt {2^{12} } = \sqrt {4096} = 64$

Also presented as
's presentation of this (of which the above is a reproduction) is not the neatest.

This is how it appears according to :


 * $64 = \surd \surd \surd \paren {4^{4!} }$