hence show that $(1-α^1)(1-α^2)(1-α^3)(1-α^4 ) = 5.$

$x^5 = 1 = cos0 + isin0$ $∴x^5= cos(2kπ) + isin(2 kπ)$ $∴x^1= (cos⁡(2kπ)+ isin(2 kπ) )^{\frac{1}{5}} = cos\frac{2kπ}{5} + isin\frac{2kπ}{5}$ Putting k =0,1,2,3,4 we get the five roots as $x_0 = cos0 + isin0 = 1 ,x_1 = cos\frac{2kπ}{5}+ isin\frac{2kπ}{5}, x_2 = cos\frac{4π}{5}+ isin\frac{4π}{5},$ $x_3 = cos\frac{6π}{5}+ isin\frac{6π}{5} ,x_4 = cos\frac{8π}{5}+ isin\frac{8π}{5}.$ Putting $x_1 = cos\frac{2π}{5}+ isin\frac{2π}{5} =αwe see that x_2=α^2, x_3=α^3, x_4=α^4$ ∴ the roots are $1, α, α^2,α^3,α^4$ and hence $∴x^5-1= (x-1)(x- αα( x-α^2 )(x- α^3 )(x-α^4)$ $∴\frac{(x^5-1 )}{((x-1) )} = (x- α)( x-α^2 )(x- α^3 )(x-α^4)$ $∴ (x- α)( x-α^2 )(x- α^3 )(x-α^4) = x^4+ x^3+x^2+x^1+1.$ Putting x=1, we get $(1-α)(1-α^2)(1-α^3)(1-α^4) = 5$