Ireland

Because $6!=720$ is not a perfect cube, we only need to consider products $t=n(n+1)(n+2)(n+3)(n+4)(n+5)$ with $n\ge 2$. Let $a=n(n+5)$, $b=(n+1)(n+4)=a+4$ and $c=(n+2)(n+3)=a+6$, then $$t=a(a+4)(a+6)=a^3+10a^2+24a.$$ Because $a\ge 14$, we have $(a+3{)}^3=a^3+9a^2+27a+27=t-a^2+3a+27=t-(a-9)(a+3)-3a<t<a^3+12a^2+48a+64=(a+4{)}^3$, hence $t$ cannot be a perfect cube.