jmc

The primes between 1 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

We compute their residues modulo 16: 2, 3, 5, 7, 11, 13, 1, 3, 7, 13, 15, 5, 9, 11, 15, 5, 11, 13, 3, 7, 9, 15, 3, 9, 1.

We multiply all of these numbers modulo 16, taking advantage of the fact that $3\cdot 5\equiv -1(\text{mod }16)$, $7\cdot 9\equiv -1(\text{mod }16)$, $11\cdot 13\equiv -1(\text{mod }16)$, and $15\equiv -1(\text{mod }16)$. We find that our answer is $\boxed{6}$.