= Solution
The complex <regular representation> has basis $(e_h)_{h\in G}$ and left action
$$
\rho(g)e_h=e_{gh}.
$$
If $\rho(g)$ is the identity, then
$$
e_g=\rho(g)e_1=e_1,
$$
so $g=1$. Thus $\mathbb CG$ is a <faithful representation>.
By <Maschke's theorem>, the regular representation decomposes as a direct sum of irreducibles,
$$
\mathbb CG=\bigoplus_jV_j.
$$
Its kernel is the intersection of the kernels of the constituent representations:
$$
\ker\rho=\bigcap_j\ker(\rho|_{V_j})=\{1\}.
$$
Each constituent kernel is a <normal subgroup> of $G$. If $G$ is simple, every such kernel is either $\{1\}$ or $G$. They cannot all be $G$, because then their intersection would be $G$, contradicting faithfulness of the regular representation. Hence some $V_j$ has trivial kernel. Therefore every finite simple group has a <faithful irreducible representation of a finite simple group>.
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