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Claim. Let $d\ge2$ and $M\ge2$ be integers with $\gcd(j^d-1:2\le j\le M)=1$. Then every integer $k\ge1+(2^d-2)(M^d-1)$ is the number of cubes in a dissection of the unit $d$-cube into homothetic cubes of positive side length. For every odd $d\ge3$ the gcd condition holds with $M=\lfloor\sqrt{3d}\rfloor+1$, so

$$c(d)\le1+(2^d-2)\bigl((\lfloor\sqrt{3d}\rfloor+1)^d-1\bigr) \qquad(d\ge3\ \text{odd}),$$

and $c(d)/d^d\to0$ as $d\to\infty$ through odd integers. Hence no constant $C>0$ gives $c(n)\ge Cn^n$ for all large $n$: the answer to whether $c(n)\gg n^n$ is no.

Covers. The definite part n_to_n of Problem 769, answered in the negative. The gcd condition is proved only for odd $d$, so no bound is given in even dimensions, including those with $d+1$ prime. The claim does not determine the order of growth of $c(n)$ or whether $\log c(n)/(n\log n)$ has a limit; the open-ended request for good bounds is not a part. The earlier pending claims of Star Fleet Math, Zeng and Korsky answer the same part the same way; none of them is a premise here.

Credit. The gcd step follows the subgroup and reduced-fraction strategy recorded in Zeng's gcd-threshold argument. Plasma AI's proof supplies every arithmetic step it needs, including its own coprime-pair estimate and the residue-path bridge from the gcd condition to the threshold. No novelty is claimed. Three earlier pending claims answer the same part the same way: Star Fleet Math's Lean disproof of 2026-07-14, Zeng's square-root threshold of 2026-07-24 and Korsky's Burgess-scale bound of 2026-08-05.

Standing. L28 is tier 0: author-recorded by Plasma AI, with a complete written proof and no independent examination or Lean proof. Catalog acceptance is pending.

Depends on. L28 (tier 0), which uses no other native claim.