Confused with imaginary calculus So $i$ is the complex unit and $n \in \mathbb{N} $.
$$e^{2 \pi \ n \ i} = 1$$
$$1^{2 \pi \ n \ i} = 1$$
$$(e^{2 \pi \ n \ i})^{2 \pi \ n \ i} = e^{-4\pi^2 \ n^2}$$
$$e^{-4\pi^2 \ n^2} \neq 1$$
I’m confused with this, can someone please explain to my where the error is in the equations above, or in my thinking? The more I sit on this problem the less I understand.
 A: Exponentiation $x^y$ is generally very tricky in complex numbers. You have to give up at least one of the following:


*

*That $x^y$ is a single-valued function.

*That $x^y$ is continuous

*That $x^y$ is defined for all $x,y, x\neq 0$.


If you keep (1), then you also have to give up $(x^y)^z = x^{yz}$ as a rule.
The best thing to do is define $x^y$ as a multivalued function. Specifically, define a multivalued $\log x$, and define $x^y = e^{y\log x}$. 
If you have a multi-valued $x^y$ then one of the values of $1^{2\pi n i}$ is $e^{-4\pi^2n^2}$.
If $y$ is rational in reduced form $\frac{p}{q}$ with $p,q\in\mathbb Z$, then $x^y$ has $q$ possible values. In particular, if $y$ is an integer, then $x^y$ has one value.
This is related to the fact that we usually pick the positive square root, but, for example, we can see $4^{1/2}=\pm 2$. In complex numbers, there are four values of $\sqrt[4]{1}=1^{1/4}$, namely, $\pm1, \pm i$.
If $y$ is not a rational number, $x^y$ has infinitely many values.
A: With complex exponentials, you aren't allowed to say $(a^x)^y=a^{(xy)}$, even for real $a$. You are just showing us a counter-example.
A: The exponential function is periodic in $\mathbb{C}$, so it's not invertible in $\mathbb{C}$. 
The confusion comes from the wrong idea that exists an inverse function  $\log x$ with domain $\mathbb{C}$.  An inverse function of exponential is well defined only if we take as his domain $\{z : -\pi<\mathcal{Im}(z)<\pi\}$ or a suitable Riemann 
surface and so we have no confusion.
In my opinion the confusion expressed in the answer is an example of how ill defined and confusing is the concept of multi-valued functions: a function must be single-value defined!
