I'm trying to prove that if $\kappa$ is a measurable cardinal with a normal ultrafilter $U$, then for every $f : [\kappa]^{< \omega} \to \gamma$, where $\gamma < \kappa$, there exists $H \in U$ homogeneous for $f$, which means that for every $n \in \omega$, $f|_{[H]^n}$ is constant.
I'm following Kanamori's book The Higher Infinite (page 83), and the proof goes like this:
If for each $n \in \omega$ there were sets $X_n \in U$ homogeneous for $f|_{[\kappa]^n}$, then $\bigcap_{n \in \omega} X_n \in U$ would be as required. Thus, it suffices to establish the following for every $n \in \omega$: for any $g : [\kappa]^n \to \gamma$, there is a set in $U$ homogeneous for $g$.
Proceeding by induction, the $n=1$ case is clear from the $\kappa$-completeness of $U$. So, assume that the assertion holds for $n \geq 1$, and suppose that $g : [\kappa]^{n+1} \to \gamma$ where $\gamma < \kappa$. For each $s \in [\kappa]^n$ define $g_s : \kappa \to \gamma$ by: $$g(\beta) = \begin{cases} g(s \cup \{\beta\}), & \text{if} \max(s) < \beta \\ 0, & \text{otherwise} \end{cases}$$ By $\kappa$-completeness, for each $s \in [\kappa]^n$ there is a $\delta_s < \gamma$ and a $Y_s \in U$ such that $g_s(Y_s) = \{\delta_s\}$. By induction hypothesis, there is a fixed $\delta < \gamma$ and a $Z \in U$ such that $s \in [Z]^n$ implies that $\delta_s = \delta$. Finally, by $\kappa$-completeness $Z_{\alpha} = \bigcap \{Y_s : \max(s) \leq \alpha\} \in U$ for each $\alpha < \kappa$, so that by normality $H = Z \cap \triangle_{\alpha < \kappa} Z_{\alpha} \in U$ (proof continues, but the rest I understood).
Now, my problem is in the bold part. I know that $Y_s \in U$ for each $s \in [\kappa]^n$, but how can I prove that $\{Y_s : \max(s) \leq \alpha\}$ is a sequence with cardinality $< \alpha$ so I can use $\kappa$-completeness? I've tried to prove by induction in $\alpha$, but the argument seems to fail when $\alpha = 0$, since $|s| \geq 2 \implies \max(s) > 0$ and then $Z_0 = \emptyset \notin U$. What did I miss?
If $\alpha<\kappa$, then there are $|\alpha|$ many finite sequences of ordinals whose maximum is at most $\alpha$.
In other words, the cardinality of $[\alpha+1]^{<\omega}$ is $|\alpha|$. But since $\alpha<\kappa$, $|\alpha|<\kappa$. So indeed by $\kappa$-completeness each $Z_\alpha$ is in $U$.