Let $N$ be the product of all odd primes less than $2^4$. What remainder does $N$ leave when divided by $2^4$?

$\text{(A) }5\qquad\text{(B) }7\qquad\text{(C) }9\qquad\text{(D) }11\qquad\text{(E) }13$
1. Identify all odd primes less than \(2^4 = 16\). These primes are \(3, 5, 7, 11, 13\).
2. Calculate the product \(N\) of these primes:
   \[
   N = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13
   \]
3. Simplify the product modulo \(2^4 = 16\):
   \[
   N \mod 16
   \]
4. Break down the product into smaller steps and use properties of congruences:
   \[
   3 \cdot 5 = 15 \equiv -1 \pmod{16}
   \]
   \[
   (-1) \cdot 7 = -7 \equiv 9 \pmod{16}
   \]
   \[
   9 \cdot 11 = 99 \equiv 3 \pmod{16}
   \]
   \[
   3 \cdot 13 = 39 \equiv 7 \pmod{16}
   \]
5. Therefore, the remainder when \(N\) is divided by \(16\) is:
   \[
   N \equiv 7 \pmod{16}
   \]

The final answer is \(\boxed{7}\).