Let $P_1,\dotsc,P_k$ be polynomials. Assume they are pairwise non-proportional (i.e., any two of them are linearly independent). Suppose $N$ is a power such that $P_1^N,\dotsc,P_k^N$ are linearly independent. Does it follow that $P_1^M,\dotsc,P_k^M$ are linearly independent for all $M \geq N$? Of course $M = N+1$ is sufficient.
The question Are large powers of polynomials linearly independent? concerns showing that there exists an $N$ giving independence. The first answer to that question shows that independence holds for all sufficiently large $N$. But a priori there might be some gaps after the first such $N$.
I am mostly interested in the case of multivariate homogeneous polynomials all of the same degree, but other cases are good too (univariate, or inhomogeneous or mixed degrees). It would be nice to have an answer over an arbitrary field, but if it helps to assume characteristic $0$ or the complex numbers then that's fine.
The answer is positive if the polynomials are homogeneous of degree $1$, i.e., linear forms. Very briefly: Let $Q_i$ be the point whose coordinates are the coefficients of the linear form $P_i$. The $P_i^M$ are linearly independent if and only if for each $i$ there's a hyperplane (in the space of degree $M$ forms) that contains all the $P_j^M$ for $j \neq i$, but not $P_i^M$; equivalently, for each $i$ there's a degree $M$ form (on the original space) that vanishes at all $Q_j$ for $j \neq i$, but not at $Q_i$; and if this holds for some degree $N$, then it also holds for $N+1$, and all $M \geq N$. (Here I am using the following identification: let $F \in \operatorname{Sym}^M(V)$. Let $L_F$ be the corresponding linear form on $\operatorname{Sym}^M(V^*)$. If we're careful about binomial coefficients, then $L_F(P_i^M) = M! F(Q_i)$.)
I really don't know the answer (let alone how to prove it). My intuition is that the answer is positive (linear independence of $N$th powers does imply linear independence of the $(N+1)$st powers) but that is only a hunch.
