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The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta N$ and $\beta N\times \beta N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.

Remark. By a result of Shelah and Velickovic, under PFA, each homeomorphism of $\omega^*=\beta\omega\setminus \omega$ is induced by some bijection between cofinite subsets of $\omega$. This implies that under PFA the space $\omega^*$ is not homeomorphic to its square.

Question. What happens under CH. Is $\omega^*$ homeomorphic to its square? The answer is affirmative if $(\omega^*)^2$ is an F-space, which means that two disjojnt open $F_\sigma$-sets in $(\omega^*)^2$ have disjoint closures.

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta\mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.

Remark. By a result of Shelah and Velickovic, under PFA, each homeomorphism of $\omega^*=\beta\omega\setminus \omega$ is induced by some bijection between cofinite subsets of $\omega$. This implies that under PFA the space $\omega^*$ is not homeomorphic to its square.

Question. What happens under CH. Is $\omega^*$ homeomorphic to its square? The answer is affirmative if $(\omega^*)^2$ is an F-space, which means that two disjojnt open $F_\sigma$-sets in $(\omega^*)^2$ have disjoint closures.

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta N$ and $\beta N\times \beta N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta N$ and $\beta N\times \beta N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.

Remark. By a result of Shelah and Velickovic, under PFA, each homeomorphism of $\omega^*=\beta\omega\setminus \omega$ is induced by some bijection between cofinite subsets of $\omega$. This implies that under PFA the space $\omega^*$ is not homeomorphic to its square.

Question. What happens under CH. Is $\omega^*$ homeomorphic to its square? The answer is affirmative if $(\omega^*)^2$ is an F-space, which means that two disjojnt open $F_\sigma$-sets in $(\omega^*)^2$ have disjoint closures.

The spaces $\beta N$ and $\beta N\times \beta N$ are not homeomorphic.

Assuming that a homeomorphism $h:\beta N\to\beta N\times \beta N$ does exist and taking into account that homeomorphisms preserve isolated points, we would conclude that $h|N$ is a bijection between $N$ and $N\times N$ and hence the unique continuous extension $\bar i:\beta(N\times N)\to\beta N\times\beta N$ of the identity map $i:N\times N\to N\times N$ is bijective. Fix any free ultrafilter $\mathcal U_0$ on $N$ and consider two ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $N\times N$. It can be shown that $\bar i(\mathcal U)=\bar i(\mathcal V)$. So, the map $\bar i$ is not injective and can not be a homeomorphism.

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta N$ and $\beta N\times \beta N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective. 

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.