So in ZFC, we have König's theorem. An important corollary is, if $\kappa$ is an infinite cardinal, then $\kappa< \kappa^ {\operatorname{cf}(\kappa)}$.
However, König's theorem is equivalent to the Axiom of Choice. So if we are working merely in ZF, I don't think we can prove $\kappa< \kappa^ {\operatorname{cf}(\kappa)}$ for infinite cardinals $\kappa$. (But correct me if I'm wrong.)
However, I was wondering if there is still an analougue of $\kappa< \kappa^ {\operatorname{cf}(\kappa)}$ for ZF if we strengthen the assumptions a little bit. Obviously it's false if $\kappa$ is infinite and $\operatorname{cf}(\kappa)$ is finite, but what if we assume both $\kappa$ and $\operatorname{cf}(\kappa)$ are infinite? Would this be strong enough, or do we need something stronger?
Edit: To clarify, the definition I'm using for $\operatorname{cf}(\kappa)$ is the least ordertype of cofinal (i.e. unbounded in $\kappa$) subsets of $\kappa$.
$\newcommand{\cof}{\operatorname{cof}}$
The way to define cofinality in the question is ill-defined for non-well orderable cardinals.
We can change a bit the definition to be: $$\cof(\kappa)=\min\{α∈Ord\mid\text{There exists a partition $P$ of $κ$ such that }|P|=α, x\in P\implies|x|<κ\}$$ For well orderable $κ$, this is the same as the usual definition, but for non-well orderable $κ$ it is more interesting:
If $κ$ is infinite Dedekind finite(wayback machine in case the link dies) then $κ$ is infinite but $\cof(κ)=2$ and indeed $κ<κ^{\cof(κ)}$
If the reals are countable union of countable sets then $\cof(|\mathbb R|)=ω$ but here we have that both $\mathbb R$ and $\cof(\mathbb R)$ are infinites and yet $|\mathbb R|=2^ω=2^{ω×ω}=(2^ω)^ω=(2^ω)^{\cof(2^ω)}=|\mathbb R|^{\cof(|\mathbb R|)}$
Indeed if the reals are countable union of countable sets then König's theorem fails but both the reals and their cofinality are infinite