jmc

Writing the product out, we get $$\frac{1\cdot 3}{5^2}\cdot \frac{2\cdot 4}{6^2}\cdot \frac{3\cdot 5}{7^2}\cdots \frac{11\cdot 13}{15^2}\cdot \frac{12\cdot 14}{16^2}\cdot \frac{13\cdot 15}{17^2}.$$The two factors of 5 in the numerators cancel the two factors of 3 in the denominators. The same occurs with the two factors of 6, and so on, up to the two factors of 13. We are left with $$\frac{2\cdot 3^2\cdot 4^2}{14\cdot 15\cdot 16^2\cdot 17^2}=\boxed{\frac{3}{161840}}.$$