Reference request: complexity of $k$-partite $k$-SAT
Claim: If there exists an $\epsilon &gt; 0$ such that for every $k'$, $k'$-partite $k'$-SAT can be solved in $2^{n(1-\epsilon)}$ time, then SETH fails.

Proof:
Suppose such an algorithm exists. We give an algorithm that, for every $k$ solves $k$-SAT in time $2^{n(1-\epsilon/2)}$.
Consider a $k$-SAT instance with $n$ variables, apply the sparsification lemma so that it produces $2^{n(\epsilon/2)}$ instances where, in every instance, every variable occurs in at most $(k/\epsilon)^{O(k/\epsilon)}$ clauses.
Each of these instances are $(k/\epsilon)^{O(k/\epsilon)}$-partite in the sense that one can find a coloring of the variables into $k' = (k/\epsilon)^{O(k/\epsilon)}$ colors such that no clause contains variables with the same color. If you want clauses of size exactly $k'$ you can replace clauses of size $k$ by $2^{k'-k}$ clauses of size $k'$.
Running the $2^{n(1-\epsilon)}$ time algorithm on each of the $2^{n(\epsilon/2)}$ instances of $k'$-partite $k'$-SAT gives an algorithm for $k$-SAT with running time $2^{n(1-\epsilon/2)}$ as claimed.

Note that the above proof essentially reduces $k$-SAT to $k^k$-partite $k^k$-SAT. So an algorithm with running time $2^{n(1-\frac{\log \log k}{\log k})}$ could still be possible, if very surprising.