Let $N$ be the product of all odd primes less than $2^4$.  What remainder does $N$ leave when divided by $2^4$?
The odd primes less than $2^4=16$ are $3,5,7,11,13$. Then \[3\cdot 5\cdot 7\cdot 11\cdot 13= (3\cdot 11)\cdot 7\cdot(5\cdot 13)=33\cdot 7\cdot 65\equiv 1\cdot 7\cdot 1 \equiv \boxed{7}\pmod {16}.\]