Problem 2.37, Fulton-Harris. Show that if $V$ is a faithful representation of $G$, i.e., $\rho: G \to GL(V)$ is injective, then any irreducible representation of $G$ is contained in some tensor power $V^{\oplus n}$$V^{\otimes n}$ of $V$.
Let $W$ be an irreducible representation of $G$, and set$$a_n = \langle \chi_W,\chi_{V^{\oplus n}}\rangle = \langle\chi_W,(\chi_V)^n\rangle.$$$$a_n = \langle \chi_W,\chi_{V^{\otimes n}}\rangle = \langle\chi_W,(\chi_V)^n\rangle.$$If we consider the generating function $f(t) = \sum_{n=1}^\infty a_nt^n$, we can evaluate it as$$f(t) = {1\over{|G|}}\sum_{n=1}^\infty \sum_{g\in G} \overline{\chi_W(g)}(\chi_V(g))^nt^n = {1\over{|G|}} \sum_{g \in G} \overline{\chi_W(g)} \sum_{n=1}^\infty (\chi_V(g)t)^n$$$$={1\over{|G|}} \sum_{g \in G}{{\overline{\chi_W(g)}\chi_V(g)t}\over{1 - \chi_V(g)t}}.$$Note that in this sum, the term where $g = e$ evaluates to $${{(\dim W \cdot \dim V)t}\over{1 - (\dim V)t}},$$which is nonzero. If no other term in the summation has denominator $1 - (\dim V)t$, then this term can not cancel, so $f(t)$ is a nontrivial rational function. We can then conclude that not all of the $a_n$ are $0$. Thus, to complete the proof, it suffices to show $\chi_V(g) = \dim V$ only for $g = e$.
