Ultrafilters: normality implying $\kappa$-completeness? In Kanamori's The Higher Infinite, it's noted at least twice (just before 17.6, and at the beginning of section 19) that if
$$\langle M,\in,U\rangle\models U\text{ is a normal ultrafilter over }\kappa$$
then $U$ is $\kappa$-complete from $M$'s perspective.  (Here $M$ is some transitive model of $\mathsf{ZFC}-\mathsf{Powerset}$, and $\kappa$ is a cardinal of $M$.)
I'm just having trouble seeing why normality would imply $\kappa$-completeness here (I don't believe it does in general), and would appreciate any help in understanding this.  Perhaps I'm missing some assumptions or something.
Explicitly, just before 17.6, it's written

For $N$ an inner model,  

$W$ is an $N$-normal ultrafilter over $\kappa$ iff
$\hspace{1em}$ (i) $W$ is an ultrafilter on $\mathcal{P}(\kappa)\cap N$, and
$\hspace{1em}$ (ii)  for any $f\in {}^\kappa \kappa \cap N$ with $\{\xi<\kappa:f(\xi)<\xi\}\in W$
$\hspace{1em}\phantom{(ii)}$ there is an $\alpha<\kappa$ such that $\{\xi<\kappa:f(\xi)=\alpha\}\in W$.

Note that (ii) subsumes what would be called $N$-$\kappa$-completeness.

Referencing this, at the beginning of section 19, Kanamori writes

Suppose that $M$ is a transitive $\in$-model of $\mathrm{ZFC}^-$; i.e. $\mathrm{ZFC}$ with the Power Set Axiom deleted.  Suppose also that $\kappa$ is an infinite cardinal in the sense of $M$... Then

$U$ is an $M$-ultrafilter over $\kappa$ iff
$\hspace{1em}$ (i) $\langle M,\in,U\rangle\models U\text{ is a normal ultrafilter over }\kappa$; and
$\hspace{1em}$ (ii) $\langle M, U\rangle$ is weakly amenable: for any $F\in {}^\kappa M\cap M$, $\{\xi<\kappa:F(\xi)\in U\}\in M$

An equivalent formulation of (i) was used for a definition before 17.6, and as noted there, normality subsumes $\kappa$-completeness in the sense of $M$, and $U\in M$ is not required.

 A: This is not true without additional hypotheses.  For instance, if for some $\lambda<\kappa$ you have a normal $\lambda$-complete ultrafilter $W$ on $\lambda$, then $W$ generates an ultrafilter $U$ on $\kappa$ which will still be normal but is not $\kappa$-complete.
It is true if you additionally assume that every element of $U$ is unbounded.  To prove it, let $\lambda<\kappa$ and let $(A_\alpha)_{\alpha<\lambda}\in M$ be a sequence of subsets of $\kappa$ that are not in $U$; we must show that the union $A=\bigcup A_\alpha$ is also not in $U$.  Define $f:\kappa\to\kappa$ by letting $f(\xi)$ be the least $\alpha$ such that $\xi\in A_\alpha$ if $\xi\in A$, and $f(\xi)=\lambda$ otherwise.  Then $f(\xi)\leq\lambda$ for all $\xi$, and so $\{\xi<\kappa:f(\xi)<\xi\}$ is cobounded and thus in $U$.  By normality, there is thus some $\alpha$ such that $\{\xi<\kappa:f(\xi)=\alpha\}\in U$.  But if $\alpha<\lambda$ then $\{\xi<\kappa:f(\xi)=\alpha\}\subseteq A_\alpha$ and thus is not in $U$.  We must thus have $\alpha=\lambda$ so $\{\xi<\kappa:f(\xi)=\lambda\}=\kappa\setminus A\in U$, as desired.
