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Yes it is easier to refute Berkeleys than Reinhardts. I learned this proof from Woodin. Assume ZFC plus $\delta$ is the least Berkeley. For each $\alpha< \delta$, fix a transitive set $M_\alpha$ containing $\delta$ such that there is no $j:M_\alpha \to M_\alpha$. Let $\gamma$ be the supremum of the ranks of the $M_\alpha$. Since $\delta$ is Berkeley there is an elementary $i : V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $i(\vec M) = \vec M$. Since $i(M_\alpha) \neq M_\alpha$, $i$ has no fixed points between its critical point and $\delta$. Therefore $\delta$ has countable cofinality: in fact, $\delta = \sup_n \kappa_n$ where $\kappa_n = j^n(\kappa)$. But then we can get another embedding $k: V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $k(\vec N) = \vec N$ where $N_n = M_{\kappa_n}$. Clearly $k$ fixes each $N_n$ since $k(n) = n$. But taking $n$ large enough that $\kappa_n > \text{crit}(k)$ yields a contradiction.