How does $e^{\pi i}$ equal $-1$

This is an interesting proof that you may have seen.

Let us define

$$e^x := \sum_{n=0}^\infty \frac{x^n}{n!}$$

Then \begin{align}e^{i \pi} &= \sum_{n=0}^\infty \frac{(i\pi)^n}{n!} \\ &= 1+i\pi-\frac{\pi^2}{2}-\frac{\pi^3}{3!} + \cdots\\ &= \left(1-\frac{\pi^2}{2}+\frac{\pi^4}{4!}+\cdots\right) + i\left(\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}+\cdots\right) \\ \\ &= \cos \pi +i\sin \pi \\ \\ &= -1\end{align}

This method can be generalized to show Euler's identity, $e^{i\theta}=\cos \theta+i\sin\theta$. From this we find that $e^z$ is actually a periodic function over $2\pi i$, and because of this, it follows that $\log z$, as the inverse of $e^z$, must be multivalued (because, $e^{i\theta} = e^{i\theta +2\pi i}=e^{i\theta -2\pi i}=\cdots = e^{i\theta+2 \pi i n}$ where $n \in \mathbb Z$).

$e^{\pi i}$ corresponds to $z = (-1, 0) = -1$ on the unit circle in the complex plane.

That is, it is positioned $\pi = 180^\circ$ from the position (1, 0) on the complex unit circle. It has no "height" in the direction of either $i$ or $-i$.

From $e^{\pi i} = -1$, we obtain Euler's Identity: $e^{\pi i} + 1 = 0$.

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Indeed one of the amazing things is that $e$ and $\pi$ have a lot to do with each other:

For example, see Euler's formula: $\quad e^{i\theta} = \cos \theta + i\sin \theta$.

So $e^{\pi i} = \cos {\pi} + i \sin {\pi} = -1 + (i \times 0)$.

A visual image may help make sense of Euler's formula and $e^{\pi i}$:

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If you really want to pursue this question, and how Euler came up with his formula, you may want take on the following challenge:

Try computing the Taylor expansion (Maclaurin Series) of $e^{\pi i}$ and see that it equals sum of the Taylor (Maclaurin) expansions of $\cos {\pi}$ and $i \sin {\pi}$. Both sum to $-1$! Or if you are lazy, you can look over at Wikipedia

BOTH $e$ and $i$ are fascinating, because they "crop up" almost everywhere in math, physics, and many other fields. For more ways of representing $e$, look at this list.

First you need to understand why, for a complex number $z$, $z = x + iy$ and $z = r e^{i \theta}$ -- where $x = r \cos \theta$ and $y = r \sin \theta$ -- are equivalent. From that, $e^{\pi i\left(1+2n\right)} = -1$ and $\ln \left(-1\right) = \pi i \left(1 + 2 n\right)$ follow for $n$ integer.

To see the former, substitute $r=1$ and $\theta = \pi\left(1+2n\right)$ into $r e^{i \theta} = r \cos \theta + i r \sin \theta$.

To see the latter, take $\ln$ of the result $e^{\pi i\left(1+2n\right)} = -1$, or take $\ln$ of $z = e^{i \theta}$: $$ \ln z = \ln r e^{i \theta} = \ln r + i \theta. $$ With $z = -1$, $r = 1$ and $\theta = \left(1+2n\right)$, and $$ \ln \left(-1\right) = \pi i\left(1+2n\right). $$