Perfect Square Product

Each $k^2-1$ can be factorised into $(k-1)(k+1)$ . Then we can just group like terms together, finding that all the terms in the middle combine to make perfect squares, leaving only $(2-1)(3-1$ ) at the front and $((n+1)-1)(n-1$ ) on the end. This can be shown algebraically with products: $\begin{aligned}(2^2-1)(3^2-1)(4^2-1)...(n^2-1)&=\prod_{k=1}^n\left(k^2-1\right)\\ &=\prod_{k=2}^n(k-1)(k+1)\\ &=\left(\prod_{k=2}^n(k-1)\right)\left(\prod_{k=2}^n(k+1)\right)\\ &=\left(\prod_{k=1}^{n-1}k\right)\left(\prod_{k=3}^{n+1}k\right)\\ &=1\times2\times\left(\prod_{k=3}^{n-1}k\right)\times n\times(n+1)\times\left(\prod_{k=3}^{n-1}k\right)\\ &=2n(n+1)\left(\prod_{k=3}^{n-1}k\right)^2\end{aligned}$ The products now always form a perfect square, so we just need to find the smallest value of $n$ such that $2n(n+1)$ is also a perfect square.

$n$ and $(n+1)$ cannot share any factors, for all factors of $n$ will leave a remainder of $1$ when divided into $(n+1)$ . This means that of $n$ and $(n+1)$ , one must be a perfect square and the other must be double a perfect square, so as to become a perfect square when multiplied by the $2$ . The first number for which this holds is $n=8$ , as $n=2\times2^2$ and $(n+1)=3^2$ .

So the answer is $\boxed{8}$ .