Proof that $g(TREE(64))<TREE(g(64))$

Let the function $g(n)$ be defined recursively such that
$<br /> g(n) = <br /> \begin{cases} <br /> g(1)= 3 \uparrow\uparrow\uparrow\uparrow 3 &amp; n = 1 \\<br /> g(n)= 3 \uparrow^{g(n-1)}3 &amp; n&gt; 1 <br /> \end{cases}<br />$

Where $g(64)$ is Graham's Number. By the Brady's Power Juice™ function $J(n)$ ,we can observe by sheer mathematical intuition, that $g(TREE(64))$ has less juice that $TREE(g(64))$

$<br /> J(g(TREE(64))) &lt; J(TREE(g(64)))<br />$

Then, by the Juice theorem, which states
that, for any two integers $m,n$

$<br /> J(n)&gt;J(m)\Rightarrow n&gt;m<br />$

We follow that

$<br /> g(TREE(64))&lt;TREE(g(64))<br />$

Q.E.D