Explain why $e^{i\pi} = -1$ to an $8^{th}$ grader? I am an $8^{th}$ grader that is taking Algebra I. But nearly everyday I try to learn things outside of what I am learning in class.     
Quite a while ago I discovered that $e^{i\pi} = -1$. This fascinated me although I do not understand it completely. I understand what imaginary and complex numbers are and what Pi is but do not completely understand Euler's number and why it has such an interesting property such as this.  
Can someone explain this to me? (Remember, I'm in $8^{th}$ grade and may not understand completely. So if someone could 'dumb it down' for me that would be very appreciated.)
EDIT: It amazes me just how much people care on this site. Each and everyone of you are putting out amazing answers (I've read them all) but, having little knowledge of trigonometry and calculus, I feel like this question is just something that will have to wait.  
Any other answers would be appreciated, but I feel as though I will not fully comprehend this until later years in school. 
Thank you, again, to everyone who made an effort to helping me understand :-).

This may not be an answer suitable to what you've learned so far, but there's a nice way to relate the $e^{i\theta}$ to sines and cosines.  When you get to calculus, you'll learn about the Taylor series representation for differentiable functions.  It turns out that for any real number $x$ you can write $\sin x$ and $\cos x$ in terms of infinite sum of numbers.  It turns out that 
\begin{eqnarray*}
\sin x & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \\
\cos x & = & 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \ldots
\end{eqnarray*}
and similarly $e^x$ has an expansion
$$
e^x \;\; =\;\; 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots
$$
If you add up $\cos x + i \sin x$ in terms of their series representation you get a nice result which is 
\begin{eqnarray*}
\cos x + i \sin x & = & 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \\
& = & 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \ldots \\
& = & e^{ix}
\end{eqnarray*}
thus giving us Euler's famous equation

$$ e^{ix} \;\; =\;\; \cos x + i\sin x$$

I understand that the preceding may be a little much for someone to understand if you don't know what series are or where they come from, but once we have Euler's equation we have a nice interpretation: If you think of the complex plane as being numbers that lie on the real axis and the imaginary axis we have that $e^{ix}$ has real component $\cos x$ and imaginary component $\sin x$, and as $x$ increases (remember that $x$ is a real number) then we trace out the arc of a circle starting at the point $(1,0)$ (notice $e^0 = \cos 0 + i \sin 0 = 1$), and this goes counterclockwise in the positive imaginary part of the plane and when we go $\pi$ radians we reach the point $e^{i\pi} = \cos \pi + i \sin \pi = -1$.
