The expression equals $\frac{n(n-1)(n-2)}{3!} \cdot \frac{n(n-1)(n-2)(n-3)}{4!} \cdot \frac{n(n-1)(n-2)(n-3)(n-4)}{5!} \cdot \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}$.

We can remove all occurrences of perfect squares (no explanation needed): so that $\frac{(n-3)(n-5)}{2 \cdot 3}=m^2$. We have $(n-4)^2-6m^2=1$. Solving the Pell equation, we get the minimal sol is $n-4=5$ and next minimal is $n-4=89$. The next minimal is $n-4=485 \implies n=489$.