Is $\omega_1 ^\omega$ countably compact? Give $\omega_1$ the order topology, and then $\omega_1 ^\omega$ the product topology.  
$\omega_1$ is countably compact, but what about this product?
I attempted to prove it in two different ways, but each time something goes wrong.  How about for arbitrary powers of $\omega_1$?
 A: It’s not hard to prove that the space $\omega_1$ is sequentially compact. The product of countably many sequentially compact spaces is sequentially compact, so $\omega_1^\omega$ is sequentially compact. Finally, in first countable spaces sequential compactness is equivalent to countable compactness, and $\omega_1$ and $\omega_1^\omega$ are first countable, so $\omega_1^\omega$ is countably compact.
Added: As Henno Brandsma reminds me, $X^\kappa$ is countable compact for every cardinal $\kappa$ iff there is a free ultrafilter $\mathscr{U}$ on $\omega$ such that $X$ is $\mathscr{U}$-compact; the result can be found (as part of Theorem $4.11$) in Jerry E. Vaughan, ‘Countably Compact and Sequentially Compact Spaces’, Handbook of Set-Theoretic Topology, K. Kunen & J.E. Vaughan, eds., North-Holland, $1984$. ($X$ is $\mathscr{U}$-compact iff each sequence in $X$ has a $\mathscr{U}$-limit, where $x$ is the $\mathscr{U}$-limit of $\langle x_n:n\in\omega\rangle$ iff $\{n\in\omega:x_n\in V\}\in\mathscr{U}$ for each nbhd $V$ of $x$.) Theorem $4.9$ of the same survey says that a space is $\mathscr{U}$-compact for all free ultrafilters $\mathscr{U}$ on $\omega$ iff it is $\omega$-bounded, meaning that the range of each sequence in $X$ is contained in some compact set. The space $\omega_1$ is clearly $\omega$-bounded, so $\omega_1$ is $\mathscr{U}$-compact for all $\mathscr{U}\in\beta\omega\setminus\omega$, and therefore $\omega_1^\kappa$ is countably compact for all $\kappa$.
A: Another idea:
It is more convient to use an equivalent definition of countable compactness. Every hausdorff space if it is countably compact iff every countably infinite subset of $X$ has an accumulation. Now we will use this definition to prove the question.
Proof: Let $A$ be a countably infinite subset of $\omega_1^\omega$. Clearly, $P_n(A)$ is countably infinite in $\omega_1^{(n)}$, where $P_n$ is the projection. Since $\omega_1^{(n)}$ is countably compact, there is an accumulation $a_n$ in $\omega_1^{(n)}$. Now let $a=\langle a_n\rangle_n \in \omega_1^\omega$. It is not difficult to see $a$ is the accumulation of $A$. This complete the proof.
