Square Product

$\begin{aligned} P & = (2^2-1)(3^2-1)(4^2-1) \cdots (n^2 - 1) \\ & = \prod_{k=2}^n (k^2-1) \\ & = \prod_{k=2}^n (k-1)(k+1) \\ & = \frac {(n-1)!(n+1)!}{2!} \\ & = \frac {\left((n-1)!\right)^2n(n+1)}2 \end{aligned}$

Note that $P$ is a perfect square when $\dfrac {n(n+1)}2$ is a perfect square. The smallest $n$ is found to be $\boxed 8$ ,