Does $\mathfrak{(p A: Andres Caicedo asked whether or not we can prove in $\sf ZF$ that there exists a counterexample (i.e. a situation where inequality holds in the assumption, but the exponents are equal).
In $\sf ZFC$ we know that there exists such counterexample, so let us assume $\sf ZF+\lnot AC$. Let $P$ be a set which cannot be well-ordered, and let $\newcommand{\fp}{\mathfrak p}\fp$ denote its cardinal. We make the following assumptions:


*

*$\fp^\omega=\fp$, otherwise replace $P$ by $P^\omega$. From this assumption we can conclude that: $\fp=\fp+\fp=\fp\cdot\fp$, and from those we can deduce that $\fp^\fp=2^\fp$.

*If $\kappa$ is the least ordinal not less or equal than $\fp$, then $\kappa<2^\fp$. If this is not true we can replace $\fp$ by $2^\fp$ (because $\kappa$ is the least ordinal with that property for $2^\fp$ as well) or by $2^{2^{\fp}}$ if needed. Since we know that $\kappa<2^{2^{2^\fp}}$, one of the three options must have the wanted property.
Note that the above properties are preserved by taking powers, so replacing $\fp$ by its power set (once or twice) would not change the first assumption.
Other important properties following from the first property are: $$2^\fp=(2^\fp)^\fp=2^\fp+2^\fp=2^\fp\cdot2^\fp.$$
We know by a lemma of Tarski that if $\lambda$ is an $\aleph$ and $\mathfrak m$ is a cardinal such that $\frak\lambda+m=\lambda\cdot m$, then $\lambda$ and $\frak m$ are comparable. Because we took $\kappa$ to be incomparable with $\fp$ we know that $\fp+\kappa<\fp\cdot\kappa$.
Note that $\fp(\fp+\kappa)=\kappa(\fp+\kappa)=(\fp+\kappa)(\fp\cdot\kappa)=\fp\cdot\kappa$ by the properties of $\fp$ and $\kappa$.


*

*Case I: $2^\kappa\nleq2^\fp$.
We observe that $2^\fp<2^{\fp+\kappa}$, and now we calculate:
$$\begin{align}
  &(2^\fp)^{\fp+\kappa}=2^{\fp(\fp+\kappa)}=2^{\fp\cdot\kappa}&\tag{1}\\
  &(2^{\fp+\kappa})^{\fp\cdot\kappa}=2^{(\fp+\kappa)\fp\cdot\kappa}=2^{\fp\cdot\kappa}&\tag{2}
  \end{align}$$
And it is not hard to see that the inequalities are satisfied, but the exponentiation ends up equal.

*Case II: $2^\kappa\leq2^\fp$.
We note that $2^\fp=2^{\fp+\kappa}$, and we make the following calculations:
$$\begin{align}
  & 2^\fp=2^{\fp+\kappa}\leq\fp^{\fp+\kappa}\leq(2^\fp)^{\fp+\kappa}=2^{\fp\cdot\kappa}=(2^\kappa)^\fp\leq(2^\fp)^\fp=2^\fp&\tag{3}\\
  & 2^\fp\leq\kappa^\fp\leq(2^\fp)^\fp=2^\fp &\tag{4}
  \end{align}$$
And in this case we see that $\kappa<2^\fp$ and that $\fp<\fp+\kappa$, but the exponentiation is again equal.
A: The inequality does not hold in general. First, note that $2^\kappa=\kappa^\kappa$ for any infinite $\kappa$. Second, it is consistent (using the technique of forcing) that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, even though $\aleph_0<\aleph_1$. This gives us that it is consistent that $\aleph_0<\aleph_1$ and yet $\aleph_0^{\aleph_0}=\aleph_1^{\aleph_1}$.
For a (combinatorially more involved) $\mathsf{ZFC}$ example, $\beth_\omega^{\aleph_0}=\beth_\omega^{\aleph_1}$, so $\beth_{\omega}^{\aleph_0}=(\beth_\omega^+)^{\aleph_1}$. (For a similar computation, see Jech's exercise 5.18, here or here.)
