Prove that: sinπ/nsin2π/n...sin(n−1)π/n=n/2^(n−1)

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`2^(n-1)sin(pi/n)sin((2pi)/n)....sin((n-1)/n)pi` equals : `2^(n-1)sin(pi/n)sin((2pi)/n)....sin((n-1)/n)pi` equals :
An aesthetically pleasing trig identity | sinpi/nsin2pi/n…sin(n-1)pi/n An aesthetically pleasing trig identity | sinpi/nsin2pi/n…sin(n-1)pi/n
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