Smallest $n$

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

For $P$ to be a perfect square, $2n(n+1)$ must be a perfect square. Since $2n(n+1)$ is even, we can assume that:

$\begin{aligned} 2n(n+1) & = (2{\color{#3D99F6}m})^2 & \small \color{#3D99F6} \text{where }m \text{ is a positive integer.} \\ n(n+1) & = 2m^2 & \small \color{#3D99F6} \text{Again, LHS is even; let }n=2p \\ 2p(2p+1) & = 2m^2 \\ p(2p+1) & = m^2 & \small \color{#3D99F6} \text{when }p=4 \\ 4(9) & = 6^2 \\ \implies p & = 4 & \small \color{#3D99F6} \text{the smallest solution.} \\ \implies n & = 2p = \boxed{8} \end{aligned}$