= Solution
Because $X$ is special, the <fundamental group of a special cube complex> embeds in a <Right-angled Artin group>. Right-angled Artin groups are residually finite by the <residual finiteness of a right-angled Artin group>, and a <subgroup of a residually finite group> is residually finite. Hence $\pi_1X$ is residually finite.
If $\pi_1X$ were simple, choose $1\ne g\in\pi_1X$. A finite quotient in which $g$ survives has a proper normal kernel. Simplicity would force that kernel to be trivial, embedding $\pi_1X$ into a finite group, contrary to the assumption that $\pi_1X$ is infinite. This is precisely the obstruction that an <infinite residually finite group is not simple>.
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