A question about relation between two fast growing sequences. Let $TREE(n)$ and $tree(n)$ are Kruskal' tree sequences. The second one is called weak.

Prove that $TREE(3)>tree^{tree^{tree^{tree^{tree^{8}(7)}(7)}(7)}(7)}(7)$

You can see that inequality in every article about $TREE(3)$ and always it is left without proof.
My real purpose is to prove that $TREE(3)>G64$ but i was only managed to prove that $tree(n)\ge 2^{n}$
Please inform me if there are simpler "combination" (with proof) of weak tree sequence so that it is still lower than $TREE(3)$ and greater than Graham's number.
 A: That is a consequence of @Deedlit's excellent answer here, which proves that  $$\text{TREE}(3)>\text{tree}_3(\text{tree}_2(\text{tree}(8)))$$
where $\text{tree}_2(n)=\text{tree}^n(n)$ and $\text{tree}_3(n)={\text{tree}_2}^n(n).$ 
Thus, letting $N=\text{tree}_2(\text{tree}(8))$, we have
$$\begin{align}\text{TREE}(3)&>\text{tree}_3(\text{tree}_2(\text{tree}(8)))=\text{tree}_3(N)\\
&>>\text{tree}^{\text{tree}^{\text{tree}^{\text{...}^{\text{tree}^{N}(N)}{...}}(N)}(N)}(N)\quad\text{with $N$ levels}\\
&>> \text{tree}^{\text{tree}^{\text{tree}^{\text{tree}^{\text{tree}^{8}(7)}(7)}(7)}(7)}(7)\quad\text{with $5$ levels}\end{align}.$$
Definitions:


*

*$\text{TREE}(n)$ is defined as the length of a longest sequence of trees $T_1,T_2,\ldots$, with vertices labelled from ${1, 2, ..., n}$ such that, for all $i,$ $T_i$ has at most $i$ vertices, and for all $i,j$ such that $i<j,$ there is no label-preserving homeomorphic embedding from $T_i$ into $T_j.$

*$\text{tree}(n)$ is defined as the length of a longest sequence of unlabelled trees $T_1,T_2,\ldots$, such that, for all $i$, $T_i$ has at most $n+i$ vertices, and for all $i,j$ with $i<j,$ $T_i$ is not homeomorphically embeddable into $T_j.$
