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.

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 $6$.
