Why isn't $e^{2\pi xi}=1$ true for all $x$? We know that $$e^{\pi i}+1=0$$and $$e^{\pi i}=-1$$
So$$(e^{\pi i})^2=(-1)^2$$$$e^{2\pi i}=1$$
Because $1$ is the multiplicative identity,$$(e^{2\pi i})^x=1^x$$$$e^{2\pi xi} =1$$should also hold true.
But we also know that $$e^{xi}=\cos(x)+i\sin(x)$$and so$$e^{2\pi xi}=\cos(2\pi x)+i\sin(2\pi x)$$which does not equal 1 for all values of $x$.
Now I realize I probably didn't break math, so I must be making an invalid assumption. What is wrong with my reasoning?
 A: 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:
$$(1-i)^{2i} \ne ((1-i)^2)^i.$$
A: In complex numbers exponentiation rules are a bit different, in this case
$$(e^{2 \pi i})^x\not\equiv e^{2 \pi i x}$$
A: Basically, you don't have the equality
$$\left(a^b\right)^c = a^{bc}$$
for complex numbers. The way I explain this to myself is that for real values, $a^b$ is defined as the limit of $a^{q_n}$ where $q_n$ is a rational number and $q_n\rightarrow b$ as $n\rightarrow\infty$. As $a^q$ is defined through roots for rational numbers, and roots become complicated for complex numbers, it makes sense that the original rule will become shaky.
A: 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.
A: When you have $e^{2\pi i x}$ you can think about the $x$ as how much you're going to rotate the number 1 (which is  $e^{2\pi i}$ in complex form) in the counter-clockwise direction. 
For example, if $x\in[0,1]$, then no rotation is just $e^0=1$, a full rotation is $e^{2\pi i \cdot 1}=1$. So for every integer-$x$, we always take $e^{2\pi i x}$ back to the real line again, where all our nice algebra rules hold.
If $x$ is not an integer (like $x=0.354$), then we have rotated 1 somewhere out in the complex plane, which means it is on the form $a+bi$. We know that $1^x$ is not on the form $a+bi$ when $x$ is a real number ($1^{0.333}=1$). So we see that clearly rule doesn't hold if $x$ is some decimal number.
Intuitively you can think that the fact that complex numbers rotate breaks the exponentiation rule, which is only true for real numbers which cannot rotate, but can only flip, e.g. $-1\cdot7$ flips $7$ to $-7$.
Edit: In fact, a full rotation is the same as two flips. So when $x$ is an integer we might as well be working with real numbers where the rule holds. [Disclaimer: maybe, look at it in any way that makes it fun.]
A: Assuming equality is true for all $x$ then $$e^{2\pi x i}=1$$ 
Using Euler's Theorem,$$\cos (2\pi x)+i\sin(2\pi x)=1$$ 
comparing real & imaginary parts on both the sides, 
$$\cos (2\pi x)=1\iff 2\pi x=2\pi k\iff x=k$$
& $$\sin(2\pi x)=0\iff 2\pi x=k\pi \iff x=\frac k 2$$
where $k$ is any integer. 
Thus, $x=k$ is the solution of given equality. The above equality will hold only & if only $x$ is an integer. Hence our assumption is wrong.   
hence 
$e^{\large 2\pi x i}=1$ is not true for all $x$ 
