I've seen stated in numerous places that if $E$ is an extender and $M$ is the ultrapower (of $V$) by $E$, then $E\not\in M$. I understand the proof of this for short extenders, but it's not clear to me how the proof should go for long extenders since the proofs I've seen all rely on the extenders being short.
What I'd like to say is something along the lines of "as $\mathrm{Ult}(V,E)$ is the direct limit of the $\mathrm{Ult}(V,E_a)$s, $E$ should appear in one of these ultrapowers but then we'd could recover $E_a\in \mathrm{Ult}(V,E_a)$ which is impossible". The issue of course is that we don't get $E_a\in\mathrm{Ult}(V,E_a)$ but merely something which, through the direct limit map $j_{a,\infty}:\mathrm{Ult}(V,E_a)\rightarrow\mathrm{Ult}(V,E)$, gets transformed into $E_a$: we get some $x\in\mathrm{Ult}(V,E_a)$ such that $E_a=j_{a,\infty}(x)$. I've tried attacking the problem in a few other ways, but have come up empty handed. Perhaps I'm missing something for such an entry-level problem.
I'm working in ZFC. Assume for now that $E$ is a set-sized extender (otherwise one has to work out what the question should mean; see below).
Let $\delta$ be the measure space of $E$, i.e. $\delta$ is an ordinal and $E$ consists of pairs $(X,a)$ such that for some $\xi<\delta$, $X\subseteq[\xi]^{<\omega}$ and $a$ is a finite set of ordinals. (E.g., $E$ is short iff $\delta=\kappa+1$ where $\kappa=\mathrm{crit}(E)$).) An ultrapower $\mathrm{Ult}(N,E)$ by $E$ is formed using functions $f:[\xi]^{<\omega}\to N$ for $\xi<\delta$.
Now suppose $E\in M=\mathrm{Ult}(V,E)$ for a contradiction. Let $i_E:V\to M$ be the ultrapower map. We will be able to compute too much of $i_E$ in $M$, by considering $\mathrm{Ult}(M,E)$ as computed in $M$.
Case 1: $\delta=\xi+1$ is a successor. So $E$ is a set of pairs $(X,a)$ where $X\subseteq\xi$. We have $\mathcal{P}(\xi)\subseteq M$, since for every subset $X$ of $\xi$, either $X$ or $\xi\backslash X$ gets into some pair in $E$. It follows that $(\xi^+)^M=\xi^+$, and that $M$ has all functions $f:[\xi]^{<\omega}\to\xi^+$ (which are all bounded in $\xi^+$, and hence coded by subsets of $\xi$). Working in $M$, let $N=\mathrm{Ult}(M,E)$ (formed using only functions in $M$) and $i^M_E:M\to N$ be the ultrapower map. Then because of the agreement between $M,V$ just mentioned, $i_E\upharpoonright(\xi^++1)=i^M_E\upharpoonright(\xi^++1)\in M$. Now $i_E$ is continuous at $\xi^+$, because all functions used to form the ultrapower are bounded in $\xi^+$. So $i_E``\xi^+$ is cofinal in $i_E(\xi^+)$. But $i_E(\xi^+)$ is regular in $M$. Therefore $i_E(\xi^+)=\xi^+$. But then $\xi^+\geq\lambda^+$ where $\lambda=\lambda(i_E)$ is the least fixed point of $i_E$ above $\kappa=\mathrm{crit}(i_E)$. Since $i_E\upharpoonright\xi^+\in M$, in particular $i_E\upharpoonright\lambda\in M$, which contradicts Kunen's theorem.
Case 2: $\delta$ is a limit. So $\mathcal{P}(\xi)\subseteq M$ for each $\xi<\delta$, and arguing as above, $i_E\upharpoonright\delta\in M$. But then $\mathcal{P}(\delta)\subseteq M$, because if $X\subseteq\delta$ then $i_E(X)\in M$, and given $\alpha<\delta$, we have $\alpha\in X$ iff $i_E(\alpha)\in i_E(X)$, and so working in $M$, using the parameters $i_E(X)$ and $i_E\upharpoonright\delta$, we can compute $X$. Now we can argue as in Case 1, but with $\delta$ here replacing $\xi$ there, for a similar contradiction.
If $E$ is a proper class extender, we can still ask whether $E$ can be amenable to $M$, i.e. $E\cap x\in M$ for all $x\in M$. But this is impossible, as it again contradicts Kunen's theorem.