TITLE: When is $(a^b)^c $ = $a^{bc}$ true?
QUESTION [8 upvotes]: I know that in some cases this rule is not true. For example
$$((-1)^2)^\frac{1}{2}\ne(-1)^{(2\cdot\frac{1}{2})}$$
So when is this rule true ?

REPLY [5 votes]: It is true when $a>0$ and $b, c \in \mathbb R$. In this situation we have a canonical choice for a value for $a^b$, given by $e^{b \ln a}$, using the principal branch of the natural logarithm on positive real numbers (that is, the one that gives $\ln a \in \mathbb R$).
For other values of $a$ there is no natural choice for $\ln a$, so $a^b$ has multiple equally valid answers. What is true is that each value of $a^{bc}$ is equal to a value for $(a^b)^c$, but not conversely. For instance, in the example you gave, $a=-1, b = 2, c = \frac{1}{2}$, the values of $a^{bc}$ are
$$(-1)^1 = e^{\ln(-1)} = e^{\pi i + 2\pi i n} = -1,$$
and the values of $(a^b)^c$ are
$$( (-1)^2)^{\frac{1}{2}} = 1^{\frac{1}{2}} = e^{\frac{1}{2} \ln(1)} = e^{\frac{1}{2} 2\pi i n} = \pm 1.$$