Showing that DNF VALID is coNP-hard
A formula $\varphi$ in DNF is valid if and only if $\neg\varphi$ in CNF is unsatisfiable. Since CNF-SAT is NP-complete, it follows that DNF-VAL is coNP-complete. You are right that you need to show an NP-complete language can be reduced in polynomial time to the complement of DNF-VAL. Since this complement is just CNF-SAT, try reducing SAT to CNF-SAT.

There are a lot of proofs of this reduction around; this one[1] looks good if you just want to see an answer.

[1] Howell, Rod. "SAT-CNF Is NP-complete." (2000).