Let $\mathcal{U}$ be a normal measure on $\kappa$ and let $j = j_\mathcal{U} : V\rightarrow M = M_\mathcal{U}$ be the corresponding ultrapower.
Prove that $\sup\{j(\alpha)\ \big|\ \alpha<\kappa^+\}$, conclude that $\{j(\alpha)\ \big|\ \alpha<\kappa^+\} \notin M$ and thus $M$ cannot be closed under $\kappa^+$-sequences of ordinals.
My try:
Let $f:\kappa\rightarrow\kappa^+$ be some function then since $\mathrm{cf}\ \kappa^+ = \kappa^+ > \kappa$ we know that $\exists\alpha<\kappa^+$ s.t
$$\forall\beta<\kappa, f(\beta) < \alpha = c_\alpha(\beta)$$
thus $[f] < j(\alpha)$. Since the above is true for every such $f$ and the ordinals below $j(\kappa^+)$ are represented by functions of the form $\kappa\rightarrow \kappa^+$ it follows that
$$\forall [f]<j(\kappa^+),\ [f]<\sup \{ j(\alpha) \big|\ \alpha<\kappa^+ \}$$
thus $\sup \{ j(\alpha) \big|\ \alpha<\kappa^+ \}=j(\kappa^+)$
What I can't figure out is the second part, why $\{ j(\alpha) \big|\ \alpha<\kappa^+ \} \notin M$?
Any help would be appriciated.