jmc

We can write the expression as $$\frac{\lfloor \sqrt[4]{1}\rfloor }{\lfloor \sqrt[4]{2}\rfloor }\cdot \frac{\lfloor \sqrt[4]{3}\rfloor }{\lfloor \sqrt[4]{4}\rfloor }\cdot \frac{\lfloor \sqrt[4]{5}\rfloor }{\lfloor \sqrt[4]{6}\rfloor }\cdots \frac{\lfloor \sqrt[4]{2015}\rfloor }{\lfloor \sqrt[4]{2016}\rfloor }.$$For each fraction, the numerator and denominator will be equal (in which case they will cancel), except when the denominator involves a perfect fourth power. Hence, the product reduces to $$\frac{\lfloor \sqrt[4]{15}\rfloor }{\lfloor \sqrt[4]{16}\rfloor }\cdot \frac{\lfloor \sqrt[4]{255}\rfloor }{\lfloor \sqrt[4]{256}\rfloor }\cdot \frac{\lfloor \sqrt[4]{1295}\rfloor }{\lfloor \sqrt[4]{1296}\rfloor }=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}=\boxed{\frac{5}{16}}.$$