Is it true that there are infinitely many $k$ such that $T(n^k)>T(n^{k+1})$?

OPEN

Let $A\subseteq \mathbb{N}$ be a complete sequence, and define the threshold of completeness $T(A)$ to be the least integer $m$ such that all $n\geq m$ are in
\[P(A) = \left\{\sum_{n\in B}n : B\subseteq A\textrm{ finite }\right\}\]
(the existence of $T(A)$ is guaranteed by completeness).

Is it true that there are infinitely many $k$ such that $T(n^k)>T(n^{k+1})$?

Erdős and Graham [ErGr80] remark that very little is known about $T(A)$ in general. It is known that
\[T(n)=1, T(n^2)=128, T(n^3)=12758,\]
\[T(n^4)=5134240,\textrm{ and }T(n^5)=67898771.\]
Erdős and Graham remark that a good candidate for the $n$ in the question are $k=2^t$ for large $t$, perhaps even $t=3$, because of the highly restricted values of $n^{2^t}$ modulo $2^{t+1}$.