Kunen showed that Reinhardt cardinals are inconsistent in ZFC. But his proof is a bit technical for a non-set-theorist to follow. Berkeley cardinals are stronger than Reinhardt cardinals. You can refute them in ZFC by observing that every Berkeley cardinal is Reinhardt, and then appealing to Kunen's theorem.
Question 1: Is there a more direct refutation of Berkeley cardinals in ZFC, perhaps one which might be more digestible by a non-set-theorist?
Recall that a definably-Berkeley cardinal is a cardinal $\kappa$ such that whenever $M$ is a transitive set with $\kappa \in M$, there are elementary embeddings $M \to M$ with arbitrarily large critical point $\alpha < \kappa$. A Reinhardt cardinal is when this holds for $M = V_\kappa$ (EDIT: In fact the "Reinhardt cardinal is defined to be the critical point of the embedding). So perhaps there is some other $M$ one can cook up which obviously fails the Berkeley property? Maybe $M = \kappa + 1$, for example?
(A Berkeley cardinal is a cardinal $\kappa$ such that whenver $M$ is a transitive set with $\kappa \in M$ and $A \subseteq M$, there are elementary embeddings $j : (M,A) \to (M,A)$ with arbitrarily large critical point below $\kappa$. Although it seems sometimes one restricts the definition to apply only when $M = V_\alpha$? I'm not sure if that's equivalent or weaker.)
Question 2: Let $\kappa$ be a cardinal (probably uncountable, regular, limit, measurable,etc.). Can it be the case in ZFC that there exists a nontrivial elementary embedding $\kappa + 1 \to \kappa + 1$?