Suppose $q:X \rightarrow Y$ is a quotient map of topological spaces such that the product map $q^2:X^2\rightarrow Y^2$ is also a quotient map. Are the maps $q^n:X^n\rightarrow Y^n$ quotient maps for all $n\geq 3$? If not, are there sufficient conditions that make this the case?

It is well-known that if $q:X\to Y$ is a quotient map, then the self-product $q^2:X^2\to Y^2$ need not be a quotient map. For instance, if $X$ is the real line generated by the basic sets $(a,b)$ and $(a,b)\backslash K$, where $K=\{1,1/2,1/3,...\}$, then the quotient map $q:X\to Y=X/K$ exhibits this failure.

Question: Suppose $q:X \rightarrow Y$ is a quotient map such that the product map $q^2:X^2\rightarrow Y^2$ is also a quotient map. Are the maps $q^n:X^n\rightarrow Y^n$ quotient maps for all $n\geq 3$?

I'd like an answer to the above question but I'd also be interested in conditions on $X$ and $Y$ which are sufficient to imply that $q^n$ is quotient for all $n$. Using a Cartesian closed category like the category of sequential (or compactly generated) spaces, it's easy to see that if $X^n$ is sequential (compactly generated), then $q^n$ is quotient if and only if $Y^n$ is sequential (compactly generated). This is interesting but I find it unfortunate that this depends on what happens to arbitrarily high powers of both $X$ and $Y$. So to clarify, I would be interested in sufficient properties which are not implied by the above fact and which do not depend on $n$.