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We know that,
z n − 1 = ( z − 1 ) ∏ r = 1 n − 1 ( z − α r )
( z − 1 ) z n − 1 = ∏ r = 1 n − 1 ( z − α r )
Applying limits on both sides,
z → 1 l im ( ( z − 1 ) z n − 1 ) = z → 1 l im ( ∏ r = 1 n − 1 ( z − α r ) )
⇒ n = ( ∏ r = 1 n − 1 ( 1 − α r ) )
= ( ∏ r = 1 n − 1 ( 1 − e n i 2 π r ) )
= ∏ r = 1 n − 1 { − e n i π r ( e n i π r − e n − i π r ) }
= ∏ r = 1 n − 1 − e n i π r . 2 i s i n ( n π r )
∣ n ∣ = ∏ r = 1 n − 1 2 s i n ( n π r )
⇒ ∣ n ∣ = 2 n − 1 ∏ r = 1 n − 1 s i n ( n π r )
∴ r = 1 ∏ n − 1 s i n ( n π r ) = 2 n − 1 n