Intuitive explanation of $(a^b)^c = a^{bc}$ What is an intuitive explanation for the rule that $(a^b)^c = a^{bc}$. I'm trying to wrap my head around it, but I can't really do it.
 A: When $b$ and $c$ are positive integers, we can make the following argument:
$$(a^b)^c=\underbrace{(a^b)\times(a^b)\times\cdots\times(a^b)}_{c\text{ times}}=$$
$$\underbrace{\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\cdots\times \left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)}_{c\text{ times}}=$$
$$\underbrace{a\times a\times \cdots\times a}_{bc\text{ times}}=a^{bc}$$
A: I will assume that $b$ and $c$ are positive integers and that $a$ is any "number" (it doesn't really matter much what $a$ is...).
Suppose I have $b \times c$ copies of the number $a$.  I can arrange them into a $b \times c$ rectangular array: i.e., with $b$ rows and $c$ columns.  When I multiply all $b \times c$ of these $a$'s together, I get $a^{bc}$.
On the other hand, suppose I look at just one column of the array.  In this column I have $b$ $a$'s, so the product of all the entries in a column is $a^b$.  But now I have $c$ columns altogether, so the product of all the entries is obtained by multiplying the common product of all the entries in a given column by itself $c$ times, or $(a^b)^c$.  
Thus $a^{bc} = (a^b)^c$.
If you want to justify this identity when $b$ and $c$ are other things besides positive integers -- e.g. real numbers, or infinite cardinals -- that's another matter: please ask.
A: If you want a physical, intuitive understanding, grab a slide rule.  As you'll be manipulating the logarithmic forms of the exponents, it should become clear quite quickly.
A: It’s not a rule in the sense of a theorem, but of a definition. (In early mathematics training, even into calculus, there is a tendency, for administrative/teaching convenience, to conflate theorems and definitions, calling all such things “rules”.)
It is simply a matter of the notation being chosen such that it is what we might call “fortuitous”. Let’s consider three other such instances.


*

*x – y = x + (-y), as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitous way in which x – y is DEFINED.

*x/y = x(1/y), for y not equal to 0, as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitious way in which x/y is DEFINED.

*a raised to the power b = exp(b*log(a)), for positive a, as you can verify from numerical examples, but it is really a matter of definition: based on extensive numerical experience, that is the fortuitous way in which mathematicians decided to DEFINE a to the power b.
Now to your question. For positive a and arbitrary b and c (or arbitrary non-zero a and integer b and c), the expression a to the power of (bc) is DEFINED, for notational convenience, to be (a to the power of b) to the power of c. The fact that it is only a notational convenience can be seen if a is negative and not both b and c are integer . Presumably you would agree with the statement that in any (valid) mathematical expression, we should be able to reduce whatever fractions occur to lowest terms without affecting the value of the expression. But that does not happen if a is negative, say, -1, b is 2, and c = 1/4: ((-1 raised to the power 2))raised to the power (1/4) is 1, and this should equal (-1 raised to the (2/4) power), which equals (-1 raised to the 1/2 power), which is the imaginary unit, not unity.
So, the short answer to your question is: It is simply a fortuitous definition, based on extensive numerical experience. Your question arose simply from the fact that you were lacking a sufficient dose of that experience. It’s not enough to just think about it: you have to grok it.
