The question is as above.
Does $e^{xy}=(e^x)^y$ hold when $x$ and $y$ are real? I remember that the answer is yes but am a little bit not confident. I know that the equality fails when $x$ an $y$ are extended to complex numbers(there is an issue of branch cut). Am I right?
Definition: for $a>0$ and any complex $b$, let $a^b$ be:
$$a^b=\exp(b\log a)$$
Therefore, for $a>0$, real $x$ and complex $y$,
$$(a^x)^y=\exp(x \log a)^y=\exp(y \log[\exp(x \log a)])=\exp(xy\log a)=a^{xy}$$
This can't apply to all complex $x$ because you need $\exp(x\log a)$ to be a positive real number to apply the definition above. And the definition does not work for complex or negative $a$ because of a branch problem with the logarithm.
Note that to write $(a^x)^y=(a^y)^x$, you need both $x$ and $y$ to be real.
Note also: $e^x$ is often an abuse of notation for $\exp(x)$, which is a function (defined on $\Bbb C$ and entire). The notation $e^x$ is "correct" because you can check easily with the definition above that $e^x=\exp(x)$. However, if you don't define the power function $a^b$ beforehand, the notation $e^x$ is meaningless.
It's a kind of circular reasoning: you would need to define $e^x$ first, in order to define $a^b$. Actually, you define $\exp(x)$ first, then $a^b$, then you can check $e^x=\exp(x)$. And that's true for every complex $x$.
But there are other ways to define $a^b$ that don't depend on the exponential. IIRC, there is something starting from Dedekind cuts: I saw this in a book once, if someone has the reference, I would really like to know!