TITLE: Consistency strength of Berkeley cardinals
QUESTION [8 upvotes]: From Cantor's Attic:

A cardinal κ is a Berkeley cardinal, if for any transitive set $M$ with $κ∈M$ and any ordinal $α<κ$ there is an elementary embedding $j : M → M$ with $\alpha<\text{crit}(j)<\kappa$. These cardinals are defined in the context of ZF set theory without the axiom of choice.

I have seen claimed several times that the existence of a Berkeley is stronger than that of a Reinhardt cardinal (let's say in NBG without Choice, as formalizing existence of Reinhardts in ZF takes some extra consideration). I can't however find any proof of this claim.
Of course if Berkeley-ness were about elementary embeddings of arbitrary classes containing $\kappa$, then it would be obvious, but we are only talking about transitive set models here, so I don't really see what to do. Can anyone provide a proof that Berkeley are stronger than Reinhardts? What about Super Reinhardts?

REPLY [7 votes]: The following is proved by Bagaria-Koellner-Woodin:
Theorem. If $δ_0$ is the least Berkeley cardinal, then there exists $γ < δ_0$ such that $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”.
See Large Cardinals beyond Choice. Let me sketch the proof, skipping some details:
First, one can show that for a tail of ordinals $\beta$, if $j: V_\beta \to V_\beta$ is an elementary embedding with $crit(j)< \delta_0$, then $j(\delta_0)=\delta_0$ and $\{\alpha < \delta_0: j(\alpha)=\alpha    \}$ is cofinal in $\delta_0$. Fix such a $\beta$ and $j$. Let $\kappa=crit(j), \lambda=κ_ω(j)$
Since $\delta_0$ is a Berkeley cardinal there are embeddings $j': V_\beta \to V_\beta$ with $\lambda <
crit(j') < \delta_0$. Also one can show $\kappa_\omega(j')< \delta_0$ holds for any such $j'$. 
So there are $\lambda'< \delta_0$ and $j': V_{\lambda'} \to V_{\lambda'}$ with $κ_ω(j')=\lambda'$ and
$crit(j') > \lambda$. Let $\lambda'$ be the least ordinal with this property and let $\gamma$
 be the least
strongly inaccessible cardinal greater than $\lambda'$. Then $j(\lambda') = \lambda'$ and $j(\gamma) = \gamma.$ 
 Thus, $j\restriction V_\gamma$
and $j'$ witness that  $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”.