Multiplicative property of exponents doesn't apply to all bases, why? Let's say we have 2 = 2^(2/2), then using the multiplicative property of exponents to turn it into a square root: ((2)^2)^(1/2) = 2, it just works. 

But in the case of -2 = (-2)^(2/2), if we use the property: ((-2)^2)^(1/2) = 2, because the ^1/2 notation means the principal root, we must pick the positive result, which is 2. 

Basically (-2)^(2/2) != ((-2)^2)^(1/2), which proves that the multiplicative property of exponents doesn't apply to all bases, what is bothering me is where this property doesn't work? For example, -2 = (-2)^(3/3) = ((-2)^3)^(1/3) = -2, it works. I have a feeling that it is when the denominator is even and the base is negative, but i can't prove it or make a logical reasoning to why and where it doesn't holds.
In general, the rule $a^{bc}=(a^b)^c$ only applies for non-negative values of $a$ (and always remember that $0^0$ is undetermined). This is because general exponentiation is usually defined via

$$a^b:= \exp(b\cdot\log(a))$$

where $\exp$ and $\log$ can be defined by e.g. power series or differential equations. The natural logarithm $\log(a)$ has a canonical value only for positive $a$, and we can extend the defnition to $a=0,b\not=0$ by setting $0^b=0$.

Writing e.g. $(-2)^{1/2}$ requires more explanation and some arbitrary choices in order to uniquely determine the value. It is dangerous to write it this way because of this ambiguity. One preferably talks about "the two solutions of $x^2=-2$" instead.



One other situation in which the rule $a^{bc}=(a^b)^c$ can be applied to all $a$, is when $b$ and $c$ are integers (still $0^0$ is excluded). That this works out well with the definition of $a^b$ given above requires some knowledge about complex numbers (I can add the reasoning if requested).
Actually the multiplicative property of exponents is valid regardless of the base but I presume  that you would require prior knowledge about complex number analysis. What you say is correct if we are dealing with numbers that are strictly real and therefore can't have imaginary or complex solutions. Hope this helps...