Why isn't $e^{2\pi xi}=1$ true for all $x$?

The notion that $(a^b)^c=a^{bc}$ has to be abandoned in complex analysis.

Or, you have to allow that $a^b$ is a multi-valued function and then you can actually say that (one of) the values of $1^x$ is $\cos(2\pi x)+i\sin(2\pi x)$. With multi-valued functions you can say "All of the values of $a^{bc}$ are values of $(a^b)^c$," but not visa versa.

Multi-valued exponentiation can be seen as an extension of the idea that there are two "square roots," and, while we usually take $\sqrt{x}$ to be the positive one, we might sometimes prefer to think of $\sqrt{x}$ as a multi-valued function. For example, if $\sqrt{x}$ is multivalued, then you can write the quadratic formula as:

$$\frac{-b+\sqrt{b^2-4ac}}{2a}$$

and no longer have that pesky $\pm$ symbol from the usual formula, being implicit in the multi-valued $\sqrt{x}$ function. But the obvious problem with multi-valued functions is that the above "looks like" it is describing a single root, when it is describing two roots.

The other problem with multivalued functions is, what would one mean by:

$$a^{b} + a^{2b}?$$Most of the time when you see something like this, you probably don't want to pick from all values of $a^{b}$ and all values of $a^{2b}$, but rather you want to pick the same "branch," which amounts to picking the same value for $\log a$ for each term, amongst the infinitely many possible values for $\log a$.

So, in short: Exponentiation in complex numbers is irritating and no fun.

The property $(a^b)^c = a^{bc}$, true in the case ${\text{positive}^\text{real}}$ (with $a^b$ always positive) isn't true in the complex case. Example from the link:

$$(1-i)^{2i} \ne ((1-i)^2)^i.$$

In complex numbers exponentiation rules are a bit different, in this case $$(e^{2 \pi i})^x\not\equiv e^{2 \pi i x}$$