good work

so cute (^^

how can a # be put to the power of i x pi AND use addition and get 0? unless youre throwing in negatives.

@Fort Zinder-Flash: that is EULER'S IDENTITY.
I know you never read this, but...

exp(i*pi)=-1, so if you add one, your result will be zero. That happens because you can write exp(i*pi) as cos(pi) i*sin(pi). (because exp(i*pi) is an immaginary number). now, sin(pi)=zero, and cos(pi)=-1, so exp(i*pi)=cos(pi) i*sin(pi)=-one zero*i=-1. omg, exp(i*pi) is -1!
Then, if you add one to exp(i*pi), you have zero :)

As mentioned by Flamiya, I will explain more clearly.
"e" is the natural number (or euler's number), ca 2,718...
"i" is the imaginary unit, a number that "i^2 = - 1"
"π" is the ratio of the lenght of a circumference and its diameter, ca 3,1415...

With the Taylor series (or better with Maclaurin series) we can prove that:

e^(iθ) = cos(θ) + i sin(θ) (Euler's formula)

where:
- "θ" is an angle (in radiant)
- "sin(θ)" is the sine function
-"cos(θ)" is the cosine function

So, for θ=π we have:

e^(iπ) = cos(π) + i sin(π) = - 1 + i * 0 = -1

And so: e^(i*π) + 1 = 0 (Euler's identity)

"We do not know what that means, but we have proved it, and therefore it must be the truth" [cit. of a famous professor after proved the Euler's identity]