smc

Combine the terms in the product to get $2^{\frac{1+3+5+\cdots +(2n-1)+(2n+1)}{7}}$. The exponent can be simplified to $$\frac{1+3+5+\cdots +(2n-1)+(2n+1)}{7}\Rightarrow \frac{\frac{n(1+(2n+1))}{2}}{7}\Rightarrow \frac{n^2}{7}.$$ We want this inequality to be true with the smallest positive odd integer value of $n$: $$2^{\frac{n^2}{7}}>1000.$$ Now, let's test the answer choices. For $n=7$, we have $2^{49/7}=2^7<1000$. For $n=9$, we have $2^{81/7}>2^{10}>1000$. So our answer is $\boxed{9}$.