 Solution

The complex regular representation has basis $(e_h{)}_{h\in G}$ and left actionIf $\rho (g)$ is the identity, thenso $g=1$. Thus $\mathbb{C}G$ is a faithful representation.

By Maschke's theorem, the regular representation decomposes as a direct sum of irreducibles,Its kernel is the intersection of the kernels of the constituent representations: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.