Why $(i-1)^{2i}\neq[(i-1)^2]^i$?

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With real numbers, we have the Power of Power Law:$$a^{xy}=(a^x)^y.$$

However, this doesn't always work with complex numbers. For example, Mathworld states that: $$(i-1)^{2i}\neq[(i-1)^2]^i.$$

I was wondering if there might be an intuitive explanation for why the Power of Power Law (sometimes) breaks down in the complex case.

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Consider how we may prove the Power of Power law for positive reals. First we must define the exponential function $$\exp x = e^x = 1+x+\frac12x^2+\frac16x^3+\cdots$$ And let $\ln x$ be its inverse (this exists because $\exp x$ is bijective). Suppose that $$e^{ab}=(e^a)^b.$$ Then $$a^{xy} = (e^{\ln a})^{xy} = e^{xy\ln a} = (e^{x\ln a})^y = (a^x)^y$$ as desired. Now, it remains to be shown that $e^{ab}=(e^a)^b$. Call the LHS $f(a,b)$ and the RHS $g(a,b)$ and treat $a$, $b$ like variables. It is easy to verify that their partial derivatives with respect to $a$ are equal, $f_a=g_a$, and similarly for that with respect to $b$, $f_b=g_b$. So these functions at most differ up to a constant, which clearly must be $0$, as $f(0,0)=g(0,0)$. Thus, the conclusion is proven.

This proof, however, breaks down when $a\in\mathbb C$. This is because the complex logarithm is infinitely multivalued, so we cannot do the manipulation we have done above.