Expressions with e, i and π

We all know that ... e^(i*π) + 1 = 0

I found out that ....

e^(i*π/3) + e^(2i*π/3) + e^(3i*π/3) + e^(4i*π/3) + e^(5i*π/3) + e^(6i*π/3) = 0

http://www.wolframalpha.com/input/?i=e^%28i*%CF%80%2F3%29+%2B+e^%282i*%CF%80%2F3%29+%2B+e^%283i*%CF%80%2F3%29+%2B+e^%284i*%CF%80%2F3%29+%2B+e^%285i*%CF%80%2F3%29+%2B+e^%286i*%CF%80%2F3%29+

And

e^(i*π/3) + 2e^(2i*π/3) + 2e^(3i*π/3) + 2e^(4i*π/3) + e^(5i*π/3) + 9e^(6i*π/3) = 6

e^(2i*π/3) + e^(4i*π/3) + e^(6i*π/3) = 0

http://www.wolframalpha.com/input/?i=e^%282i*%CF%80%2F3%29+%2B+e^%284i*%CF%80%2F3%29+%2B+e^%286i*%CF%80%2F3%29+

e^(3i*π/3) + e^(6i*π/3) = 0

http://www.wolframalpha.com/input/?i=e^%283i*%CF%80%2F3%29+%2B+e^%286i*%CF%80%2F3%29

6e^(6i*π/3) = 6

http://www.wolframalpha.com/input/?i=6e^%286i*%CF%80%2F3%29+

6*[cos(2π) + i*sin(2π)] = 6

http://www.wolframalpha.com/input/?i=cos%282%CF%80%29+%2B+i*sin%282%CF%80%29+

6/π^2

http://www.wolframalpha.com/input/?i=6%2F%CF%80^2+

It's also the probability that a number picked at random from the set of integers will have no repeated prime divisors

http://primes.utm.edu/curios/page.php?short=6

e^(2i*π) = e^(4*i*π) = e^(6*i*π) = 1
e^(3i*π) = -1 .... e^(5i*π) = -1 .... e^(7i*π) = -1

e^(2!*i*π) = e^(3!*i*π) = e^(4!*i*π) = e^(5!*i*π) = 1