Let $\hat P(A)=P(A)-\{\emptyset\}$, and $A_0=\mathbb{N}$ and $A_{n+1}=\hat P(A_n)$.
I was asked to prove $|A_n|=|A_n\times A_n|$ for any $n\in\mathbb{N}$.
I thought about using induction: We have a bijection $\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ so the statement holds for $A_0$.
Assume it holds for all $k < n$, we have :
$$|A_n|=|\hat P(A_{n-1}) \times \hat P(A_{n-1})|\space \space definition$$ And $$|A_{n-1}|=|A_{n-1} \times A_{n-1}| \space \space inductive \space assumption$$ So : $$|P(A_{n-1})\times P(A_{n-1})|=|A_{n-1}\to\{0,1\}\times A_{n-1}\to\{0,1\}|=2^{|A_{n-1}|}2^{|A_{n-1}|}=2^{2|A_{n-1}|}=^{(*)}2^{|A_{n-1}|}=|P(A_{n-1})|$$ (*) - by the induction assumption
So we finally have : $$|A_n|=|\hat P(A_{n-1}) \times \hat P(A_{n-1})|=|A_n\times A_n|=$$
Now, the next question requires to find where the proof of this statement fails for $P(A)$ instead of $\hat P(A)$, but my proof did not use the fact we are removing the empty set anywhere, So I`m pretty sure I have a mistake.
I'd be happy if someone can correct me and offer alternative proofs.