Nothing is wrong with Mathematica with respect to the statements you appear resistant to acknowledging that those aren't generally true (despite counterexamples given to you). I would argue that is is you who needs to supply the assumptions appropriate for your objectives.

And you've used the wrong form and location for Assuming in the code you posted. (An indication that an error occurred should have been observed.) The statement in your code

a^p/a^q == a^(p - q), Assuming[a, a != 0]

should probably be

Assuming[b > 0 && a > 0 && p > 0, (a/b)^p == a^p/b^p]

although that doesn't get you anywhere. You can explore with various assumptions. Consider the following:

FullSimplify[(a/b)^p == a^p/b^p, Assumptions -> {b > 0}]
(* True *)
FullSimplify[(a/b)^p == a^p/b^p, Assumptions -> {b < 0 && a < 0}]
(* True *)

While I am no mathematician, it does appear that the truth of your equality depends on various assumptions.

Ok, but is that a bug in Mathematica or suggest an improved rules or how to provide the correct assumptions to those rules. I am only following what Mathematica gives me. Please help me to do better, thanks.

Thank you all for sharing my wrestling with Mma stating the basics, at least for now on Roots. I put a notebook together to study. Suggestions and corrections welcomed.

I understand once you move into typeset statements, it can get dicey how Mma understands the statements; FullForm versions are often the best way.

The ref nb on PowerExpand needs to have examples of using it in proofs/proving.

One of my fav teachers told me, you should back to re-study the basics every 5 or 10 yrs. ;-) I still think mma should have understood my truth testing without having a deep knowledge of other functions.

Oh! back to the negative raised.... Thats what I started out with. So what Mathematica gave via copy as plaintext and pasted was correct all along? Suggestions on a workflow going forward?

Power[Power[a,Power[q,-1]],Power[p,-1]] == Power[a,Power[pq,-1]]

well, it still gives False even with PowerExpand. I guess only visually symbolically I state its true. Is it my poor understanding using Mathematica or Mathematica can't really do it at this time? Or need to load up on a long list of assumptions?

(a^p)^(1/q)//FullForm
(* Power[Power[a, p], Power[q, -1]] *)

What did you do to get back Power[a^p, q^-1]? Its standard form is (a^p)^(1/q).

Here is a counterexample:

With[{p = 2, q = 3, a = -1},
 a^(p/q) == Power[a^p, q^-1]]

I always tell my students that it is very dangerous to elevate a negative basis to a non-integer exponent, because the familiar algebraic rules break down. Is seems to be a little-known issue.

Then I tried this:

Assuming[
 p > 0 && p \[Element] Reals && a > 0 && a \[Element] Reals && q > 0 &&
   q \[Element] Reals,
 a^(p/q) === Power[a^p, q^-1]]

Nope; then I tried this:

  SameQ[a^(p/q),Power[a^p,q^-1], Assumptions->{a>0,p>0,q>0}]

Nope again. And it appears SameQ does not understand Assumptions (Assuming ref nb). Should I had to have read PowerExpand or Simplify before proving that the comparison is true? Am asking a larger question for those expecting Mathematica to just get it right without the deep understanding of how Mathematica works. Translating from a standard textbook on math appears much harder than I realized.

Right. Understand better. So testing for SameQ requires spelling out the assumptions to prove the lhs===rhs. But from my reading SameQ does not use Assuming/Assumptions; or did I misunderstand.

I and others have wanted better tools for proofs and proving within Mathematica for a long time. I would have expected by now the pattern some of the high school math states be parsed better in Mathematica. PowerExpand was out of left field for proving; there are no examples of using it for proving. Learned a cool insight. :-)

Cool, and very instructive, thank you; much to study. However, I was expecting the kernel to understand the original statement without the deep contextualizing used. I wanted to take a simple high school math relationship and confirm its truth within Mathematica. Note:

WolframAlpha["a^(p/q)==Power[a^p,q^-1]", \
{{"AlternateFormAssumingAllVariablesAreRealAndPositive", 1}, 
  "Plaintext"}]

Tells me the relationship is True as I would have expected.

PowerExpand works.