 Solution

Let the distinct values of the faithful character $\chi$ be $a_1,\dots ,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi ,\dots ,{\chi }^{r-1}$, thenfor $0\le n<r$. The Vandermonde matrix is invertible, so every $S_j=0$. But faithfulness implies that $\chi (g)=\chi (1)$ only for $g=1$: in a unitary realization, equality of the trace with the dimension forces every eigenvalue to be one. For that fiber $S_j=\psi (1)>0$, a contradiction.