Answer: $n\geq 3$ We show by induction that the proposition $P_n: (2^n+1)!!-1\equiv 0\pmod{2^{n+1}}$ is true for $n\geq 3$. $n=3$ is obvious. Assume we have shown $P_k$ for any $k \leq n$. Then, $$ (2^{n+1}+1)!!-1\equiv (2^{n+1}+1)\prod_{k=1,k: odd}^{2^n-1} (2^n-k)(2^n+k)\equiv (2^{n+1}+1)\prod_{k=1,k: odd}^{2^n-1} (2^{2n}-k^2)\equiv (2^{n+1}+1) (2^n-1)!!^2. \pmod{2^{n+2}} (1) $$ From $P_n$, $(2^n+1)\cdot (2^n-1)!! = 2^{n+1}t + 1$. For some integer $t$. Squaring both side and taking $\pmod{2^{n+2}}$, $$ (2^n+1)^2\cdot (2^n-1)!!^2 = \left(2^{n+1}t + 1\right)^2 $$ or $$ (2^{2n}+2^{n+1}+1)\cdot (2^n-1)!!^2 = 2^{2(n+1)}t^2+2^{n+2}t + 1 $$or $$ (2^{n+1}+1)\cdot (2^n-1)!!^2 \equiv 1. $$Thus, the R.H.S of (1) is equivalent to 1, which means $P_{n+1}$ is true, completing induction. Note that it is easy to check that $n=2$ is invalid.