Excellent! Just to make it even more explicit that a^(p/q) == Power[a^p, q^-1] is not generally true using your example:

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

With[{p = 2, q = 3, a = -1}, a^(p/q)] // ComplexExpand
(* -(1/2) + (I Sqrt[3])/2 *)

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

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.

Beware, your algebraic rules break down easily:

In[2]:= With[{a = -I, b = I, p = 1/2}, (a/b)^p == a^p/b^p ]

Out[2]= False

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.

pq is not the same as p q. You forgot to put a space inbetween.

(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).

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. :-)

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.

yep,

PowerExpand[Power[Power[a, q^-1], p^-1] == Power[a, (p q)^-1]]

returns True

PowerExpand works.