This question arises from a proof in Kanamori's "The Higher Infinite". To get started, assume that $a^\#$ exists for every $a \in \omega^\omega$. Also for any $a \in \omega^\omega$, let $I_a$ denote the corresponding club of indiscernibles that generate $L[a]$. And also note that for any $a$, $I_{a^\#} \subseteq I_a$.
Now the question is that what happens to successor elements of $I_a$ in $L[a^\#]$. And the claim is that if $\langle \xi_\alpha: \alpha \in \text{On}\rangle$ is the monotone enumeration of $I_a$, then in $L[a^\#]$ for any $\alpha$, $|\xi_{\alpha}| = |\xi_{\alpha+1}|$.
Note. This claim is used for $\xi_\alpha \lt \aleph_\omega$, but since there was no emphasis on it, I think it holds in general.
I have thus far showed that for any $\alpha$, $\xi_{\alpha+1} \not \in I_{a^\#}$. But this is not enough as $L[a^\#]$ has many cardinals that are not indiscernible.
I would appreciate any hints or solutions for this.
Minor side note. You can assume the background theory to be ZF + AD + DC.
The existence of $a^{\#}$ implies that every uncountable cardinal is a limit member of the club of $a$-indiscernibles. If you apply that fact within the model $L[a^{\#}]$, you'll get the result you wanted.