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.