$M\models 2^\alpha=\alpha^+\ \forall \alpha<\kappa\longrightarrow 2^\kappa=\kappa^+$ if $M$ is a model of ZFC given by a measurable cardinal
In Jech 1978, page 319 (lemma 28.13), we read:
Let $\kappa$ be a measurable cardinal. If $2^{\kappa} > \kappa^+$, then the set $\{\alpha < \kappa : 2^α > α^+\}$ has measure one for every normal measure on κ. Consequently, if $2^α = α^+$ for all cardinals $α < κ$, then $2^κ = κ^+$.
Proof. Let $D$ be a normal measure on $κ$, and let $M = Ult_D(V )$. If $2^α = α^+$ for almost all $α$, then, since $[d]_D = κ$, we have $M \models 2^κ = κ^+$.
Now why is the latter true? I proved that $[d]^+=[d^+]$, where $d$ is the identity function of $\kappa$ seen in the ultrapower and $d^+$ is the function $d(\alpha)=\alpha^+$ for all cardinals $\alpha<\kappa$ (and arbitrarily defined on the other ordinals less than $\kappa$, since they have measure $0$). But now I don't know how to use this to conclude.
Thank you in advance for any suggestions.
EDIT: Sorry for my previous question,the comouter saved it wrong.
The fact that $M \models 2^{\kappa} = \kappa^{+}$ follows immediately from Łoś's theorem. In fact $$ \{ \alpha < \kappa \mid 2^{\alpha} = \alpha^+ \} = \{ \alpha < \kappa \mid \operatorname{GCH}_{d(\alpha)} \} \in D $$ and thus $$ M \models \operatorname{GCH}_{[d]}, $$ i.e. $$ M \models 2^{\kappa} = \kappa^{+}. $$
However, we also get:
Claim. $V \models 2^{\kappa} = \kappa^{+}$.
Proof. Since $M \subseteq V$, we have that $(\kappa^{+})^{M} \ge (\kappa^{+})^{V}$ and, since $V$ and $M$ have the same powerset of $\kappa$, we get that $V \models 2^{\kappa} \le (\kappa^{+})^{M}$. It hence suffices to show that $(\kappa^{+})^{V} = (\kappa^{+})^{M}$.
Suppose otherwise. Then $(\kappa^{+})^{V} < (\kappa^{+})^{M}$ and thus, in $M$, there is a well-order $R \subseteq \kappa \times \kappa$ of order type $(\kappa^{+})^{V}$. This set exists in $V$ which is absurd. Q.E.D.