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.

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.
