Prove: for. Euler's equation: From which we can get: So each of the factors in the original expression is: where k goes from 1 to n-1 Factor out the first term: The original expression becomes a product of these three things: #1. #2. #3. where k goes from 1 to n-1 Adding the exponents in #1. #1. The sum of the first n-1 integers is , so the exponent of e becomes and #1 is now #1. Since #3 it gives the product: Putting #1 and #3 together: #1*#3 Now we look at #2 again. #2. Those factors are all (1 - an nth root of 1 other than 1 itself. To show that 1 = cis(0 + 2k*pi) = e^(2k*pi) and by DeMoivre's theorem the n nth roots of unity are cis(2k*pi/n) and that's what the second terms in those parenthetical expressions are. and we can use any n-1 consecutive even integers for k, so here we are using the n-1 negative consecutive even integers -2, -4, -6, -2(n-1). And they won't include 1 itself because we are not including integers 0 or 2n. The nth roots of 1 can also be gotten by solving the equation The first parentheses give us the root of one which is 1 itself. The n-1 degree polynomial in the second parentheses has solutions which are all the n-1 nth roots of 1 other than 1. Therefore the polynomial #4. x^(n-1)+x^(n-2)+x^(n-3)+"..."+1 is equivalent to a polynomial which is like #2 with x's placed where the 1's are. #5. because they have the same solution and both have leading coefficient 1. So when we substitute 1 for x in #5, we get #2 and when we substitute 1 for x in #4 we get n because there are n terms. Therefore #2 simplifies to just n Therefore #1*#2*#3 = Edwin