Explain why $e^{i\pi} = -1$ to an $8^{th}$ grader?
Instead of proving it, I'll just explain how we can represent any complex number in polar form.
Any complex number can be represented in Cartesian form (the form you're probably familiar with) as $x+iy$, where $x,y \in \Bbb R$ (that means that $x$ and $y$ are real numbers) and $i^2 = -1$.
However all complex numbers can be represented in another form as well: polar form. Polar form involves specifying the distance from the origin and the angle that the ray through the point makes with the positive $x$ axis. We can represent complex numbers by $re^{i\theta}$ where $r$ is the distance, $\theta$ is the angle, and $i$ is the same as the one above.
To convert between the two, you use Euler's formula $$re^{i\theta}=r(\cos(\theta) +i\sin(\theta)) = x+iy$$
Here's an image that might help:
Now you can see for yourself that $e^{i\pi} = 1e^{i\pi}$ has a length of $1$ and makes an angle of $\pi$. So it lies along the $x$-axis, $1$ unit to the left of the origin: i.e. it is $-1$.
If you graph $$y = e^x$$ and $$ y = 1 + x, $$ the graphs look sort of similar near $x = 0$. If instead of $ 1 + x$ you graph $y = 1 + x + \frac{x}{2!}$, it looks even better. And if you graph $$ y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^{10}}{10!}, $$ it looks even better. If you think about a polynomial with infinitely many terms, following the same pattern, you can (with some real effort!) make sense of it, sort of the same way you make sense of $1 + (1/2) + (1/4) + \ldots = 2$. And that infinite polynomial will have a graph that looks exactly like the graph of $y = e^x$. So people say that $e^x$ and that infinite polynomial are "the same thing." (We call the polynomial a "power series" representation, because it involves successive powers of $x$, and it's an infinite series.)
Once you've said that $e^x$ and the power series are the same thing, it makes sense to talk about $e^z$, where $z$ is complex: you just use the polynomial to evaluate it.
You can do the same kind of thing for $\sin$, getting $$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \pm \ldots $$ and $$ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4}! \pm \ldots $$
[By the way... the way you guess these polynomials is to use calculus which isn't really what you were hoping for, I suppose. But among all first-degree polynomials, $y = 1 + x$ is the one that best fits $y = e^x$ near $x = 0$: the value at $0$ is correct, and the slope at zero is also correct. The higher-degree representations similarly are "best fits", but in more complicated ways.]
Once you've done that for sine and cosine, you can do some algebra to show that $$ e^{ix} = \cos x + i \sin x $$ although you need to rearrange terms a lot, and proving that this is valid for these particular infinite polynomials is remarkably subtle.
But from that last great formula, you can plug in $x = \pi$ to get Euler's formula.
As in my other answer, I won't be trying to prove anything to you -- power series, differential equations, etc. will be over your head for the moment. Instead I'll offer a different interpretation of the exponential function that hopefully you will find enlightening.
Instead of seeing $e^{i\theta}$ as a number -- let's interpret it as an operator on complex numbers. Think of it as a generalization of the negation operator "$-$". Geometrically, what negation does is it keeps the length of the arrow pointing from the origin to a number on the real number line the same, while flipping its direction.
On the real number line there are only two different directions from the origin. So negation is a pretty easy thing to pin down. Not so with complex numbers. Graphing complex numbers requires us to plot on a plane. So from the origin there are an infinite number of different directions.
So, not only could one potentially flip the direction of the arrow pointing to a complex number -- one could rotate it. That's what the exponential function $e^{i\theta}$ does -- it rotates a complex number about the origin by an angle $\theta$ (measured in radians, not degrees) without changing its length.
Let's look at an example. We'll start with the complex number $z=1 + i$. That one looks nice and easy. Now let's say that we want to rotate $z$ counterclockwise by an angle of $\pi /6$ to get a new complex number $z'$. Then we just use the exponential operator:
$$z' = e^{i\pi/6}z = e^{i\pi/6}(1+i)$$
Now actually finding an explicit form for $z'$ will require some geometry and trigonometry -- but it shouldn't be too hard. Let's give it a try.
First we'll graph $z$:
From the Pythagorean theorem, we can see that the hypotenuse of that triangle is $\sqrt{1^1+1^2} = \sqrt{2}$. And from the fact that $$\cos(\theta) = \dfrac{\text{adjacent}}{\text{hypotenuse}}$$ we see that $\theta = \operatorname{arccos}\left(\frac 1{\sqrt{2}}\right) = \pi/4$.
Now we want to rotate that arrow (the hypotenuse of the triangle) that we're using to point to our complex number by $\pi/6$ while keeping its length the same. Well we can immediately see the angle that we'll get -- it'll just be $\pi/4 + \pi/6 = 5\pi/12$. So our new complex number $z'$ should have a picture that looks like this:
Now using some trig, we see that the base of that triangle should be $\sqrt{2}\cos\left(\dfrac {5\pi}{12}\right) = -\dfrac 12 + \dfrac {\sqrt{3}}2$ and the height should be $\sqrt{2}\sin\left(\dfrac {5\pi}{12}\right) = \dfrac 12 + \dfrac {\sqrt{3}}2$.
Therefore $$z' = e^{i\pi/6}(1+i) = \left(-\dfrac 12 + \dfrac {\sqrt{3}}2\right) + \left(\dfrac 12 + \dfrac {\sqrt{3}}2\right)i$$
Now that we see how it works, let's consider the formula $$e^{i\pi} = -1$$ It looks like we can't use our approach -- afterall what is $e^{i\pi}$ operating on? What's it's operating on is the complex number $1=1+0i$. That is $e^{i\pi}$ is exactly the same as $e^{i\pi}(1)$.
So now it should be clear what we need to do -- we need to rotate the complex number $1$ by the angle $\pi$ radians. But $\pi$ radians is just halfway around a full circle. So $e^{i\pi}(1)=-1$.
