My question is specifically about $\emptyset$, but more generally about any language that can be decided in (deterministic or nondeterministic doesn't really make a difference here) constant time. Obviously, $\emptyset$ cannot be $\mathsf{NP}$-hard since there is nothing to map "yes"-instances to, but I'd prefer a one line argument using a time hierarchy theorem, ideally. Currently I'm arguing (very informally speaking) that since constant time is $\subsetneq \mathbb{o}(n \log n)$, constant time is strictly "less powerful" than any deterministic polynomial time due to the deterministic time hierarchy theorem and thus - since $$\mathsf{Polytime}(f(n)) \subseteq \mathsf{NPolytime}(f(n))$$ - also strictly less powerful than nondeterministic polynomial time. So, constant time languages can't be $\mathsf{NP}$-hard.
This doesn't sound elegant to me at all (although it should be correct I hope, even if not phrased as formally as I'd write it down for something to hand in). What's the most elegant argument you can come up with, preferably using not-too-advanced complexity theory?