The converse of Tychonoff theorem? 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.
 A: If $A$ contains at least two points and $\kappa$ is infinite, then it is easy to see that every open set in $A^\kappa$ is infinite; in particular $A^\kappa$ has no isolated points.  But of course we can construct a compact space of cardinality $2^\kappa$ that does have isolated points; for instance the disjoint union of $\{0,1\}^\kappa$ (with its product topology) and one additional point.  Therefore such a space cannot be homeomorphic to $A^\kappa$ for any $A$.
Regarding the case $X = [0,1]$: it is not homeomorphic to any product $A^\kappa$.  Suppose it were.  Since $A$ is the continuous image of $X$ under the coordinate maps, $A$ is compact and connected.  $A$ is also homeomorphic to a subset of $X$ under any map like $x \mapsto (x, x_0, x_0, \dots)$.  But the only compact connected subsets of $[0,1]$ are: the empty set, singletons, and closed intervals.  The first two cases are absurd, and it's easy to see that $[0,1]$ is not homeomorphic to $[0,1]^\kappa$ for any $\kappa > 1$, since $[0,1]^\kappa$ remains connected when any one point is deleted.
I suppose one might ask for necessary and sufficient conditions for a compact space $X$ to be homeomorphic to a product.  I don't know what they might be.
A: The answer is no. Very simple counter examples can be constructed as follows. Consider the cardinal $4=2^2$. Every finite topological space is compact. There are four topologies on a set with two elements. That means that there are at most $4$ spaces of cardinality $4$ that are of the form $A^2$ for some $2$-point space $A$. However, there are 355 distinct topologies on a four point set. So, certainly many many $4$-set topologies are not of the form $A^2$ for any $2$-set topology $A$. (Note that relaxing the question here to $X$ being a product up to homeomorphism or even homotopy is still answer negatively by a similar counting argument.)
For infinite counterexamples, here is an algebraic topology argument. It is well known that for any finitely generated group $G$ there exists a compact CW-space $X$ whose fundamental group is $G$. The fundamental group functor respects arbitrary products: $\pi_1(\prod Xi)\cong \prod\pi_1(X_i)$. So, now just take $G$ to be any non-trivial simple group. If the corresponding $X$ would be a product of $\kappa$ many spaces than $G$ would be a non-trivial product and thus not simple. 
A: 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 $Z$.
