2016 gone wrong

$(2016)^{2017} \equiv 16^{2017} \equiv 16^{\big(5 \times\phi(10^3) + 17\big) } \equiv 2^{4\times 17} \pmod{10^3}$

Now comes the messy part.

$2^7 \equiv 3 \pmod{5^3}$

$2^{65} = 2^2\cdot (2^{7})^{9} \equiv 4(3^{9}) \equiv 4(3^{5})(3^{4}) \equiv 4(-7)(81) \equiv 107 \pmod{5^3}$

Now we multiply the whole congruence with $2^3$ .

$2^{65+3} \equiv 107\times 8 = 856 \pmod{5^3 \times 2^3}$

Hence, the last $3$ digits of $\displaystyle (2016)^{2017}$ are 856.

Their product when divided by $320$ gives $\boxed{240}$ .