Is there a model of $\mathsf{ZFC}$ such that for every cardinal $\beta > \aleph\_0$ there is a cardinal $\alpha < \beta$ such that $|{\cal P}(\alpha)|
>\beta$?



Model in $\mathsf{ZFC}$ such that ${\cal P}(\ldots)$ has "jumping property"
There cannot be such a model. Let for instance $\beta=\beth\_\omega$, the supremum of $\beth\_0=\aleph\_0,\beth\_1=2^{\beth\_0},\beth\_2=2^{\beth\_1},\dots$, for any $\alpha<\beta$ we have $\alpha\leq\beth\_n$ for some $n$, and hence $|\mathcal P(\alpha)|\leq\beth\_{n+1}<\beta$.


The exact same argument shows that this proposition necessarily fails for all $\beta$ which are strong limit cardinals (i.e. $\kappa<\beta\implies 2^\kappa<\beta$). However, at least relative to large cardinals, it is consistent that those are the only points of failure - by a result of Woodin (apparently unpublished, discussed [here](https://mathoverflow.net/q/138308/30186)) it is consistent with ZFC that for all $\alpha$ we have $2^{\aleph\_\alpha}=\aleph\_{\alpha+2}$. This implies that the strong limit cardinals are precisely $\aleph\_\alpha$ with $\alpha$ limit. Hence if we take any infinite cardinal which is not a strong limit, it is of the form $\aleph\_{\beta+1}$. Then we have $\aleph\_\beta<\aleph\_{\beta+1}$ but $2^{\aleph\_\beta}>\aleph\_{\beta+1}$.


