= Solution
The map $a\mapsto\{a\}$ injects $x$ into $\mathcal P(x)$. Cantor's diagonal argument shows that no map $x\to\mathcal P(x)$ is surjective: for $f:x\to\mathcal P(x)$, the set $D=\{a:a\notin f(a)\}$ differs from every $f(a)$. Hence $\kappa<2^\kappa$.
If $f:x\to y$ is surjective, inverse image gives an injection
$$
\mathcal P(y)\to\mathcal P(x),
\qquad A\mapsto f^{-1}(A).
$$
For every infinite initial ordinal $\omega_\alpha$, transfinite recursion constructs a pairing of $\omega_\alpha\times\omega_\alpha$ with $\omega_\alpha$: at each stage fewer than $\aleph_\alpha$ earlier pairs have been used. Thus
$$
\aleph_\alpha\aleph_\alpha=\aleph_\alpha
$$
without invoking choice for arbitrary families.
Every ordinal below $\omega_{\alpha+1}$ has cardinal at most $\aleph_\alpha$ and can be coded by a well-ordering relation on a subset of $\omega_\alpha$. Sending such a relation to its order type, and all non-well-orders to zero, gives a surjection
$$
\mathcal P(\omega_\alpha\times\omega_\alpha)\to\omega_{\alpha+1}.
$$
Using the pairing above to code relations as subsets of $\omega_\alpha$ yields
$$
\boxed{\aleph_{\alpha+1}\leq2^{\aleph_\alpha}}.
$$
Again, strictness beyond this is independent of the usual axioms.
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