How to show $\alpha^n$ is star countable for any ordinal $\alpha$?
A topological space $X$ is said to be star countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.
Thanks ahead.
I claim that $\alpha^n$ has countable extent. If $\operatorname{cf}\alpha\le\omega$, then $\alpha^n$ is $\sigma$-compact and hence has countable extent. If $\operatorname{cf}\alpha>\omega$, let $$\sigma=\left\langle\langle\xi_k^1,\dots,\xi_k^n\rangle:k\in\omega\right\rangle$$ be a sequence of distinct points in $\alpha^n$. Then $\sigma$ has an infinite subsequence $$\sigma_1=\left\langle\langle\xi_{n_k}^1,\dots,\xi_{n_k}^n\rangle:k\in\omega\right\rangle$$ such that each of the sequences $\sigma_i=\langle\xi_{n_k}^i:k\in\omega\rangle$ for $i=1,\dots,n$ is either constant or strictly increasing, and at least one is strictly increasing. If $\sigma_i$ is constant, let $\xi^i=\xi_{n_0}^i$, and if $\sigma_i$ is not constant, let $\xi^i=\lim\,\sigma_i$; then $\langle\xi^1,\dots,\xi^n\rangle$ is a limit point of $\left\{\langle\xi_k^1,\dots,\xi_k^n\rangle:k\in\omega\right\}$. Thus, $\alpha^n$ does not contain an infinite closed discrete set and therefore has countable extent.
The result now follows from the fact that a space with countable extent is star countable. Let $X$ be a space with countable extent, and let $\mathscr{U}$ be an open cover of $X$. If $\eta$ is an ordinal, and $x_\xi\in X$ have been chosen for $\xi<\eta$ so that $x_\xi\notin\operatorname{St}\big(\{x_\zeta:\zeta<\xi\},\mathscr{U}\big)$ for each $\xi<\eta$, let $A_\eta=\{x_\xi:\xi<\eta\}$, and if $\operatorname{St}(A_\eta,\mathscr{U})\ne X$, choose $x_\eta\in X\setminus\operatorname{St}(A_\eta,\mathscr{U})$. At some point we must have $\operatorname{St}(A_\eta,\mathscr{U})=X$, and $A_\eta$ is then a closed, discrete subset of $X$. Since $X$ has countable extent, $\eta<\omega_1$, and $X$ is star countable.