The title says it, basically. My question is $-$ for $ n \ge 2 $, can $n!!$ be a perfect square, where $!!$ represents the double-factorial? My conjecture is no, but I can't seem to be able to find a good proof for this.
No, for odd $n,$ there is a prime between $(n-1)/2$ and $n.$ The exponent of this prime in factoring $n!!$ is one, that is, odd.
Edit. looking at even numbers and the definition, it appears $$ (2n)!! = 2^n n! $$ in which case we ignore the exponent of $2$ and concentrate on $n!,$ which cannot be a square either for this $n \geq 3,$ also because of an odd prime. See here.