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If `A=((p,q),(0,1))`, then show that `A^(8)=((p^(8),q((p^(8)-1)/(p-1))),(0,1))`

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`A^(2)=((p,q),(0,1))((p,q),(0,1))`
`=((p^(2),pq+q),(0,1))=((p^(2), q(p+1)),(0,1))`
`A^(3)=A.A^(2)`
`=((p,q),(0,1))((p^(2), pq+q),(0,1))=((p^(3),p^(2)q+pq+q),(0,1))=((p^(3), q(p^(2)+p+1)),(0,1))`
Similarly,
`A^(4)=((p^(4), q(p^(3)+p^(2)+p+1)),(0,1))` and so on.
`:. A^(8)=((p^(8), q(p^(7)+p^(8)+...+1)),(0,1))=((p^(8),q((p^(8)-1)/(p-1))),(0,1))`

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