TITLE: The converse of Tychonoff theorem?
QUESTION [5 upvotes]: Suppose, a topological space  $(X, \mathscr{T})$ is compact, and the cardinality of $X$ is $2^\kappa$. Is there a compact space $A$ with the cardinality not larger than $2^\kappa$, such that $A^\kappa$ with the product topology $\mathscr{T'}$coincide with $(X, \mathscr{T})$? In other words, Could $(X, \mathscr{T})$ be homeomorphic to $(A^\kappa, \mathscr{T'})$?
Added: As pointed out in Ittay Weiss's and Nate Eldredge's answers, the answer is negative when $\kappa$ is finite and when  $(X, \mathscr{T})$ contains a isolated point. I become interested in the condition that guarentees the homeomorphism exists. In particular, I'm interested in the case when $\kappa$ is infinite, and no isolated point in  $(X, \mathscr{T})$, say, the closed interval with the usual topology.

REPLY [2 votes]: Assume that $X$ is infinite. If $X$ is homeomorphic to $Y^\kappa$ for some infinite $\kappa$, then $X\cong Y^\omega\times Y^\kappa$, and $Y^\omega$ has cardinality at least $2^\omega$. For each $y\in Y^\omega$ let $X_y=\{y\}\times Y^\kappa$; then $X_y\cong X$, so $\{X_y:y\in Y^\omega\}$ is a partition of $X$ into $|Y|^\omega$ closed subsets homeomorphic to $X$.
Now let $\kappa$ be any infinite cardinal. Yet $Y=\{0,1\}^\kappa$ with the product topology, where $\{0,1\}$ is discrete, let $D_\kappa$ be a discrete space of cardinality $2^\kappa$, let $X=D_\kappa\times Y$, and let $X^*$ be the one-point compactification of $X$. Then $X^*$ is a compact Hausdorff space of cardinality $2^\kappa$ without isolated points, and the point at infinity is the unique point of $X^*$ whose character (i.e., minimum cardinality of a local base) is $2^\kappa$, so $X^*$ does not contain even two subspaces homeomorphic to itself. In particular, $X$ is not homeomorphic to $Z^\kappa$ for any space $Z$.