Firstly, the smallest beautiful number equals $3\times 4\times 5=60$. Therefore the answer cannot be greater than $60$. We are going to show that the answer is $\boxed{60}$. This is equivalent to showing that $60\mid (n^2-1)n^2(n^2 + 1)$ holds for any positive integer $n \ge 2$.

Write $(n^2-1)n^2(n^2+1)$ as $\Big((n-1)n(n+1)\Big)\Big(n(n^2+1)\Big)$. Its first part, $(n-1)n(n+1)$ is a product of three positive integers, therefore, it must be a multiple of $3!$. For the second part $n(n^2+1)$, either $n$ or $(n^2+1)$ will be even. Therefore it is a multiple of $2$. Together, the whole thing must be a multiple of $3! \times 2=12$.

Next, we are going to show it is also a multiple of $5$. By the conclusion of # 4156, $n^2\equiv 0, \pm 1\pmod{5}$. Therefore, one of $(n^2-1)$, $n^2$, and $(n^2+1)$ must be a multiple of $5$.

Therefore, $(n^2-1)n^2(n^2+1)$ must be a multiple of $12\times 5=60$. This means the greatest common divisor of all beautiful numbers is $60$.