I think the problem is wrong $1.ω^{ 2 }.ω^{ 3 }.ω^{ 4 }......ω^{ 71 }=ω^{ n }$

$n=$ $\sum _{ x=2 }^{ 71 }{ (x) } =2555$ and 2555 isn't a multiple of 3

$\therefore$ $1.ω^{ 2 }.ω^{ 3 }.ω^{ 4 }......ω^{ 71 }=ω^{ 2555 }=ω^{ 2 }$

The problem should be $ω.ω^{ 2 }.ω^{ 3 }.ω^{ 4 }......ω^{ 71 }$

then the answer is $\boxed { 1 }$