 Solution

Let $K=\ker \chi$, which is nontrivial. Every tensor power affording ${\chi }^n$ is trivial on $K$, so each of its irreducible constituents has $K$ in its kernel. Some irreducible representation is nontrivial on $K$; otherwise the regular representation, which contains every irreducible, would also be trivial on $K$. Choose its character $\psi$. Then $\psi$ is a constituent of no ${\chi }^n$, and