Proof of index laws for complex numbers Can someone give a proof that index laws (and hence log laws) apply for complex numbers in the same way they do to reals, specifically that:

$(a^{ix})^n = a^{ixn}$

Assuming $a, x, n$ are real and $i$ defined as $i^2 = -1$
I am trying to prove de Moivre's theorem from Euler's formula but have been told I can't just assume that index laws apply to complex numbers (but rather have to prove that they do).
 
Help would be greatly appreciated!

Thank you!!
You have to be careful about multivalued functions.  By definition, $a^z = \exp(z \log(a))$, but $\log$ is multivalued: $\log(a) = \text{Log}(a) + 2 \pi i k$ where $\text{Log}$ is one particular branch of the logarithm, and $k$ can be any integer. 
Now 
$(a^{ix})^n = (\exp(i x \log(a))^n = \exp(n \log(\exp(i x \log(a)))$
and $\log(\exp(i x \log(a))) = i x \log(a) + 2 \pi i k$ for arbitrary integer $k$
so 
$$(a^{ix})^n = \exp(n (i x \log(a) + 2 \pi i k)) = a^{ixn} \exp(2 \pi i k n)$$

If $n$ is an integer, so is $k n$, so $\exp(2 \pi i k n) = 1$ and you do have
$(a^{ix})^n = a^{ixn}$ (i.e. every value of the left side is a value of the right side, and vice versa).
However, if $n$ is not an integer they can be different.  

For example, $(i^2)^{1/2} = (-1)^{1/2} = \pm i$, but $i^{2 \cdot 1/2} = i$.
Using logarithms (in base a) we get $(a^{ix})^n \to n\log_a{a^{ix}}=ixn$ which is the same as the right hand side.