How is this sorting algorithm Θ(n³) and not Θ(n²), worst-case?
This algorithm can be re-written like this


Scan A until you find an inversion.
If you find one, swap and start over.
If there is none, terminate.


Now there can be at most $\binom{n}{2} \in \Theta(n^2)$ inversions and you need a linear-time scan to find each -- so the worst-case running time is $\Theta(n^3)$. A beautiful teaching example as it trips up the pattern-matching approach many succumb to!

Nota bene: One has to be a little careful: some inversions appear early, some late, so it is not per se trivial that the costs add up as claimed (for the lower bound). You also need to observe that swaps never introduce new inversions. A more detailed analysis of the case with the inversely sorted array will then yield something like the quadratic case of Gauss' formula.

As @gnasher729 aptly comments, it's easy to see the worst-case running time is $\Omega(n^3)$ by analyzing the running time when sorting the input $[1, 2, \dots, n, 2n, 2n-1, \dots, n+1]$ (though this input is probably not the worst case).

Be careful: don't assume that a reversely-sorted array will necessarily be the worst-case input for all sorting algorithms.  That depends on the algorithm.  There are some sorting algorithms where a reversely-sorted array isn't the worst case, and might even be close to the best case.