= Solution
Let the distinct values of the faithful character $\chi$ be $a_1,\ldots,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi,\ldots,\chi^{r-1}$, then
$$
0=|G|\langle\chi^n,\psi\rangle
=\sum_{j=1}^ra_j^nS_j,
\qquad
S_j=\sum_{\chi(g)=a_j}\overline{\psi(g)}
$$
for $0\leq n<r$. The <Vandermonde determinant>[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.
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