Why $e^{i(π/3)} \ne -1$? I understand why $e^{i\pi} = -1$and as a result  $e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$ These results can be confirmed using Euler's formula
But why does $e^{i\pi/3}\neq -1$ as we can write it $(e^{i\pi})^{1/3}$
and as such $(-1)^{1/3}$ the cubic root of $-1$ is $-1$ itself.
Yet using Euler's formula we get $e^{i\pi/3}=\frac{1}{2}+i\frac{\sqrt 3}{2}$.
 A: In the complex case a number not equal to $0$ has three cubic roots.
A: The problematic step is when you say "the cubic root of $-1$ is $-1$ itself". Yes, $(-1)^3 = -1$, but there are other cube roots of $-1$, namely $e^{i(\pi/3)}$, for example. The problem is that $(-1)^{1/3}$ is multi-valued. 
A: I'm surprised that none of the other answers fail to point out the actual error in your question (no it is not that $-1$ has $3$ cube-roots):

It is not at all true that $(x^a)^b = x^{ab}$ for complex numbers $x,a,b$ in general, under any reasonable definition of complex exponentiation.

Counter-examples (for various conventions):

$-1 = (-1)^3 = (-1)^{6/2} \color{red}{\large\boldsymbol\ne} ((-1)^6)^{1/2} = 1^{1/2} = 1$
$i = (-1)^{1/2} = (-1)^{2/4} \color{red}{\large\boldsymbol\ne} ((-1)^2)^{1/4} = 1^{1/4} = 1$

It is wrong to think you can easily solve this using multi-valued functions. Even if $1^{1/4}$ is the set $\{1,i,-1,-i\}$, it is still not the same as just $i$.
You must be absolutely clear about the conditions under which certain identities hold. Relevant are the following:

$x^{ab} = (x^a)^b$ for any real number $x$ and natural numbers $a,b$.

If you want the more general fact for integer exponents:

$x^{ab} = (x^a)^b$ for any real number $x \ne 0$ and integers $a,b$.

In fact it turns out that 'miraculously' we have an even more general fact for real exponents:

$x^{ab} = (x^a)^b$ for any real number $x > 0$ and reals $a,b$.

Notice that all these precise statements about real exponentiation show you clearly that you must know exactly what the objects are before you can apply any operations to them, not to say claim any properties about the resulting values.
In particular, the above identities are simply meaningless if you do not specify what $x,a,b$ are!
For this reason it is actually an important question to ask whether there are corresponding rules for complex numbers.
Yes, but not as nice.

$x^{ab} = (x^a)^b$ for any complex number $x \ne 0$ and integers $a,b$. (*)

Here exponentiation is simply the result of starting from $1$ and repeatedly multiplying/dividing by $x$ where the number of times is specified by the exponent (multiplying for positive; dividing for negative). This fact holds in any structure that has invertible multiplication, including the field of rationals, the field of reals, and the field of complex numbers.
$x^{ab},x^a$ are well-defined since $x \ne 0$.
However, in general "$x^{ab} = (x^a)^b$" does not hold for complex $x$ even if $a,b$ are both rational, as already shown above.
So it's excellent that you ask whether some new structure (complex numbers) have the same properties as some other structure (real numbers) instead of just blindly assuming it does.

In general (*) holds for any field.
Suppose we have a field $S$ (such as the complex numbers) and an exponentiation operation that satisfies the following:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

$x^0 = 1$ for every $x \in S$.
$x^{k+1} = x^k x$ for every $x \in S$ and integer $k$.

Note that any reasonable foundational system is capable of defining such an operation recursively (you need one direction for positive $k$ and another for negative $k$), and can easily prove by induction the following two theorems.


$x^{a+b} = x^a x^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.
Then $x^{a+0} = x^a = x^a x^0$.
Given any integer $b$ such that $x^{a+b} = x^a x^b$:
  $x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = ( x^a x^b ) x = x^a ( x^b x ) = x^a x^{b+1}$.
  $x^{a+(b-1)} = x^{(a+b)-1} = x^{a+b} \div x = ( x^a x^b ) \div x = x^a ( x^b \div x ) = x^a x^{b-1}$.
Therefore by induction $x^{a+b} = x^a x^b$ for every integer $b$.

$x^{ab} = (x^a)^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.
Then $x^{a \times 0} = x^0 = 1 = (x^a)^0$.
Given any integer $b$ such that $x^{ab} = (x^a)^b$:
  $x^{a(b+1)} = x^{ab+a} = x^{ab} x^a = (x^a)^b (x^a) = (x^a)^{b+1}$.
  $x^{a(b-1)} = x^{ab-a} = x^{ab} \div x^a = (x^a)^b \div (x^a) = (x^a)^{b-1}$.
Therefore by induction $x^{ab} = (x^a)^b$ for every integer $b$.

Notice that we did not use commutativity here, which in fact shows that the argument holds in any division ring. If you restrict the exponents to natural numbers, then it clearly holds in any group when "nonzero" is deleted.
