If $$ e^{ix} = \cos{x} + i \sin{x} $$ which means $$ e^{2\pi i} = \cos{2\pi} + i \sin{2\pi} = 1 + 0i = 1 $$ and $$ a^{(bc)} = (a^b)^c $$ then why is this wrong for some real number $x$? $$ e^{ix} = e^{2\pi i(x/2\pi)} \\ = (e^{2\pi i})^{(x/2\pi)} \\ = 1^{(x/2\pi)} \\ = 1 $$ ?
I am sure there is something wrong i am doing but i just want someone to point it to me.  Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$ Why do we have to calculate it by the formula as it is (which produces different value)  Another question is why can't we say that $ (-1)^{\frac {1}{2}} = \left(\left (-1\right)^{2}\right)^{\frac {1}{4}} = 1^{\frac {1}{4}} = 1$  Again i have to say that i know this is all wrong i just want to know why it is wrong
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