$T_4$-ness that is preserved by product Sorgenfrey line demonstrates how normality can be not preserved when "squared."
Is there an example for a normal space $X$ for each of?:


*

*$X^2$ is normal, but $X^3$ is not

*$X^2$ and $X^3$ are normal, but $X^4$ is not

*$X^k$ is normal for $2 \leq k < n$, but $X^n$ is not

*$X^n$ is normal for $n \in \mathbb N$, but $X^\omega$ is not

*$X^n$ is normal for $n \in \mathbb N$, but $X^\omega$ is not, in box topology
 A: For the first four questions, the answer is "yes", and I believe was first established in the paper


*

*Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.


Therein the following theorem is proved:

Theorem 1.1.  For every $k$ and $m$ such that $1 \leq k \leq m \leq \omega$ there exists a separable and first-countable space $X = X ( k , m)$ such that
  
  
*
  
*$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$;
  
*$X^n$ is normal (collectionwise normal) if and only if $n < m$.
  

The answer to the last question is also "yes", and an example was essentially established in 


*

*van Douwen, Eric K., Another nonnormal box product, General Topology Appl. 7, 71-76 (1977). ZBL0341.54008.



Theorem B. $\Box^\omega ( 2^{\omega_2} )$ is not normal. 

(I.e., the box product of $\omega$ copies of the Tychonoff product $2^{\omega_2} = \{ 0,1 \}^{\omega_2}$ is not normal.) Clearly $2^{\omega_2}$ is compact Hausdorff, hence normal, and each finite Tychonoff product $( 2^{\omega_2} )^n$ is homeomorphic to $2^{\omega_2}$.
