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Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answerthis answer.

Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answer.

Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answer.

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David E Speyer
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Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$.

Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then $$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$ We want to show that this Hom space is nontrivial for large $N$.

We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of $(*)$ is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.

I explain how to modify this for compact Lie groups in this answer.