This was a homework problem for a course that I am a TA for. The solution that I had in mind involved using a Vandermonde determinant argument (See Theorem 19.10 in the book by James and Liebeck). But, I was surprised by the following beautiful solution that was submitted by multiple students:
Let $V$ be a faithful representation and $W$ an irreducible representation of $G$. Let $a=dim(V)$$a=\dim(V)$ and $b=dim(W)$$b=\dim(W)$, and let their respective characters be $\chi$ and $\psi$. Then, for all $g\in G$, we have $|\psi(g)|\leq b$, whereas, due to faithfulness, for all $g\in G\setminus\{e\}$, we have $|\chi(g)|\leq a-\epsilon$$|\chi(g)|\leq a-\varepsilon$ for some $\epsilon>0$$\varepsilon>0$. Then, we have:
\begin{align*} |\langle\chi^n,\psi\rangle|&=\frac{1}{|G|}|\sum_{g}\chi(g)^n\overline{\psi(g )}|\\ &\geq \frac{1}{|G|}(a^nb - \sum_{g\neq e}|\chi(g)^n\overline{\psi(g)}| )\\ &\geq \frac{1}{|G|}(a^nb-(|G|-1)(a-\epsilon)^nb), \end{align*}\begin{align*} |\langle\chi^n,\psi\rangle|&=\frac{1}{|G|}\left|\sum_{g}\chi(g)^n\overline{\psi(g )}\,\right|\\ &\geq \frac{1}{|G|}\left(a^nb - \sum_{g\neq e}\left|\,\chi(g)^n\overline{\psi(g)}\,\right| \right)\\ &\geq \frac{1}{|G|}\big(a^nb-(|G|-1)(a-\varepsilon)^nb\big), \end{align*} and as $n\rightarrow \infty$, the above expression becomes positive, showing that the inner product of $\psi$ is non-zero with some power of $\chi$, and thus, $W$ is a sub-representation of some tensor power of $V$, completing the proof!
