If F is the field, write FG for the group algebra and call V the FG-module associated with a given representation of G. For any non zero v in V, Gv has at most |G| elements, and so the vector subspace W = span(Gv) has dimension at most |G| and it is clearly stable under the action of G (i.e., it is an FG-submodule of V). But W is non trivial and so if V is irreducible, it must be that W=V. Thus |G|>=dim(W)=dim(V).
I think this work, but according to Dummit & Foote Exercice 5 in the section on representation theory, we can do better and show that an irreducible representation has dimension strictly less than |G|!