Why doesn't $[a]^{[b]} = [a^b]$? When we construct the modular numbers $\Bbb Z/n\Bbb Z$ so that addition works as $[a]+[b]=[a+b]$ and multiplication works as $[a][b]=[ab]$, we get for free that $a[b]=[ab]$ also works.  Why can't we also define a sense of exponentiation $[a]^{[b]}$ as $[a^b]$?  We do have that $[a]^b = [a^b]$ because $$[a]^b = [a]\cdots [a] = [a\cdots a] = [a^b]$$ but what is different about exponentiation vs addition and multiplication that we can't define $[a]^{[b]} = [a^b]$?
Example where the formula doesn't work: Let our modular system be $\Bbb Z/4\Bbb Z$, then $$[5]=[1], \text{ but} \\ [2]^{[5]} = [2^5] = [0] \ne [2] = [2^1] = [2]^{[1]}$$
I'm looking for an intuitive explanation as to why this doesn't work right when the similar formulas for $+$ and $\cdot$ do.
 A: The key notion here is a congruence — "congruent modulo $n$" was constructed to be an equivalence relation that respects addition and multiplication, but need not preserve anything not constructed from them.
So, it's not really about addition/multiplication being special at all — it's about the relation being constructed.
As an example consider instead the rational numbers modulo $1$; that is, $p \equiv q$ if and only if $p-q$ is an integer. This is a congruence for addition: $[p+q]=[p'+q]$ whenever $[p] = [p']$, but it can't be such for multiplication: e.g. $[1/2] = [3/2]$, but $[1/2 \cdot 1/2] \not\equiv [3/2 \cdot 1/2]$.
Similarly, we could consider the equivalence relation on the integers defined by $x \equiv y$ if and only if either $x$ and $y$ are both zero or $xy$ is a nonzero square. This equivalence relation is a congruence for multiplication: $[xy]=[x'y]$ whenever $[x]=[x']$, but it can't possibly respect addition: $[1] \equiv [4]$, but $[1+1] \not\equiv [1+4]$

A note on language: normally the kind of structure is implicit, and we just speak of congruences. e.g. if we're talking about rings, then we simply say "congruence relation" for an equivalence relation that respects addition, negation, and multiplication.
A: Arithmetic $\mod n$ is determined by the divisibility of classes and the resulting remainders.   An exponential power will not in any way be divided by nor have any remainder in terms of $n$.
Example:  $15^6$ means we multiply $15$ by itself $6$ times.  $3|15$ is important and so $3|15^6$.  That $3|6$ is completely irrelevant.  We are not doing any arithmetic operation ON the $6$. The $6$ only tells us how many times we are doing an operation on something else.  The $6$ is not being acted open in any way.
So $15^6 \mod 7 \equiv (2*7 + 1)^6 \mod 7 \equiv 1 \mod 7$ but nowhere in the equation do we have any arithmetic activity happening to the $6$.
A: $$[a]^{[b]} \overset{?}{=} [a^b]$$
It's not uncommon to assume properties about a certain notation that aren't true. For example, $\sqrt{a+b} \ne \sqrt a + \sqrt b$. I've always found that plugging in numbers and observing the results is helpful.
It is true that
$$[13^{15}] = [3]^{15}= [3]^{5} \cdot [3]^{10}$$
If this is going to equal $$[13]^{[15]} = [3]^{[5]}$$
then we are going to have to require that 
$$[3]^{10} = [1]\; \text{instead of}\; [9]$$
one more try
It isn't hard to show that $[a^b] = [a]^b$. So you could rephrase your question as

Why isn't $[a]^b = [a]^{[b]}?$

If the domain is $\mathbb Z_B$, then there exists a positive integer, $\beta$, such that $b = [b] + \beta B$ where $0 \le [b] < B$.
Then $[a]^b = [a]^{[b]+\beta B} = [a]^{[b]} \cdot [a]^{\beta B}$
So, in order to have $[a]^b = [a]^{[b]}$, it must be true that $[a]^{\beta B} = 1$. Since $\beta$ is basically arbitrary, we really must have $[a]^B = 1$. This is usually not true.
A: You might like Euler's theorem. It says that when $a$ and $n$ are coprime, we have $a^{\varphi(n)}\equiv_n1$. Taking your bracket notation and adding a subscript to indicate which modular group we're talking about, this means that $[a^b]_n=[a]_n^{[b]_{\varphi(n)}}$ (again, whenever $a$ and $n$ are coprime). So, we can do modular arithmetic on exponents in many cases -- it just isn't necessarily modular with respect to $n$!
This certainly doesn't show that arithmetic modulo $n$ on the exponent definitely can't work, but it can be treated as a clue:
there's no reason to believe that $\varphi(n)$ is a factor of $n$, or that just because arithmetic mod $\varphi(n)$ works therefore arithmetic on the greatest common divisor of $\varphi(n)$ and $n$ works.
