Sorry to resurrect such an old thread, but we supply yet one more proof. We follow the notation of Exercise 2.37 in Fulton-Harris, which asks the same question as this thread.

Let $W$ be an irreducible representation of $G$, and set$$a_n = \langle \chi_W,\chi_{V^{\oplus n}}\rangle = \langle\chi_W,(\chi_V)^n\rangle.$$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 $g = e$.

Suppose $\chi_V(g) = \dim V = n$ for $g \neq e$. Also, say $G$ acts on $V$ via $\rho: G \to GL(V)$. There is $k$ such that $\rho(g)^k = I$. If $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $g$ we have$$\lambda_1^{ik} + \dots + \lambda_n^{ik} = n$$for $i = 0, 1, \dots$. Since $g^{k+1} = g$, we also see$$\lambda_1^{ik+1} + \dots + \lambda_n^{ik+1} = n.$$It follows that $$\lambda_1^{ik}(\lambda_1 - 1) + \dots + \lambda_n^{ik}(\lambda_n - 1) = 0$$which implies for all polynomials in $\mathbb{C}[x]$, we have$$P(\lambda_1^k)(\lambda_1 - 1) + \dots + P(\lambda_n^k)(\lambda_n - 1) = 0.$$Choosing appropriate polynomials with roots at all but one of the eigenvalues, we see that all the eigenvalues must be $1$. Since $\rho(g)$ is diagonalizable, it follows $\rho(g) = I$. This contradicts the faithfulness of $V$.