Proving that $e^\pi=e^{-\pi}$ I've been stuck with this for a while now. I have this chain of reasoning that would imply $e^{-\pi}=e^\pi$, obviously false, since $e^\pi$ and $e^{-\pi}$ are two real distinct numbers and so I must have made an assumption somewhere that I cannot actually do. I know I have to be very careful when working with complex numbers, especially when they're in the exponents, and so I tried to make the steps as small as I could so that it would be easier to point out where it went wrong.
\begin{align}
e^{-\pi}&= e^{\pi\cdot -1}\tag{1}\\
&=e^{\pi\cdot i^2}\tag{2}\\
&= e^{\pi\cdot i\cdot i}\tag{3}\\
&= \left(e^{\pi\cdot i}\right)^i\tag{4}\\
&= (-1)^i\tag{5}\\
&=\left(\tfrac{1}{-1}\right)^i\tag{6}\\
&=\left((-1)^{-1}\right)^i\tag{7}\\
&=(-1)^{-1\cdot i}\tag{8}\\
&=(-1)^{-i}\tag{9}\\
&=\left(e^{i\pi}\right)^{-i}\tag{10}\\
&=e^{i\pi\cdot -i}\tag{11}\\
&=e^{-i^2\pi}\tag{12}\\
&=e^\pi\tag{13}
\end{align}
I suspect it has something to do with changing the base from $e$ to $-1$, but what does that mean? Are complex powers only defined for positive bases? Any help is appreciated.

Here is your "proof" presented differently:
We have $e^{i\pi}=-1=\frac{1}{-1}=\frac{1}{e^{i\pi}}=e^{-i\pi}$. So far everything is right. Now our idea is to take both sides to the power of $i$: $(e^{i\pi})^i=(e^{-i\pi})^i$. The erroneous conclusion would appear if you used the identity $(a^b)^c=a^{bc}$. And here lies the problem: this identity doesn't hold for all complex numbers. (EDIT: in fact, this identity isn't always true if we have real numbers $a,b,c$ as leonbloy mentions in the comment. Keep that in mind!)
One might also touch on the topic: What is $(e^{i\pi})^i$? Here we need to go back to definition of exponentiation of complex numbers: $a^b=e^{b\ln a}$. However, there is a serious problem here: complex logarithm is multivalued. Taking one branch of the logarithm, we have $\ln e^{i\pi}=i\pi$, so $(e^{i\pi})^i=e^{i\cdot i\pi}$ (NOTE: here we use the definition of complex exponentiation, not exactly the property $(a^b)^c=a^{bc}$), which is $e^{-\pi}$.
However, at the same time we have $e^{i\pi}=e^{-i\pi}$, so we could say $\ln e^{i\pi}=-i\pi$. That way we get $(e^{i\pi})^i=e^{i\cdot (-i\pi)}=e^\pi$.
So if you think about this for a while, the core of the problem is that complex logarithm, and hence also exponentiation, are multivalued.
