Question: I was recently taught exponentials and I decided to play around with negative bases, which they told me were not allowed. The obvious place to start was negative one, and, as expected, the graphing tool did not work. However, after trying Wolfram, it told me it was equaled $e^{i\pi x}.$ I would understand why it oscillates, like the sine and cosine waves (negative base to an even power becomes positive, negative to an odd power is negative) but why are they exactly the same?

Answer: 
Because one logarithm of $-1$ is indeed $i \pi,$ as in $$ e^{i \pi} +1 = 0. $$

Another logarithm is $3 i \pi,$ so a less common assignment for $(-1)^x$ would be $$ e^{3 i \pi x} $$

One application that is not widely known is with the Gelfond-Schneider Theorem, using your expression. Since $-1$ is not $0$ or $1$ and is algebraic (indeed rational and an integer), **if $x$ is irrational and algebraic** over $\mathbb Q,$ 
then $$ e^{ i \pi x} = \cos \pi x + i \sin \pi x $$
is **transcendental**. In turn, that means the real numbers $\cos \pi x$ and $\sin \pi x$ are transcendental.

I used that in an article. 

Really.
Usually for $a$ positive, you define

$$a^x := e^{x \ln(a)}$$

Now, for $a$ negativve, $\ln(a)$ is not defined. You can still work around that by noticing that

$$-1 = e^{i\pi}$$

So, it would makes sense to define 

$$\ln(-1) := i\pi$$

The problem is that it makes as much sense to define it as 

$$\ln(-1) := i(2k+1)\pi$$

So, you could define

$$(-1)^x := e^{x i(2k+1)\pi }$$

And by convention, you **choose** $k=0$