Let $A$ be an infinite set. Then $|A|=|A|^n$ for any $n\in \mathbb{N}$.
What I would like to do is adapt a proof of the Comparison Lemma. I've tried to do it in the following way:
Let $n=2$ (and then proceed by induction, once it has been proved that $|A| = |A|^2$). First of all, it is trivial that $|A|\le |A\times A|$. Now I need to show that $|A\times A| \le |A|$.
Let $\Delta = \{(B,f): B\in \mathcal{P}(A)\setminus \{\emptyset\}, f:B \to A\times A \mbox{ is injective } \}$. Then $\Delta\ne \emptyset$, since for $b\in B, f: b\to (b,b)$ exists.
Let $(B,f)\preceq(C,g)\iff B\subseteq C \subseteq A, g_{\vert B}=f$, where $f:B\to A^2, g:C\to A^2$. Let $\Gamma = \{B_i, f_i\}_{i\in I}$ be a chain in $(\Delta, \preceq)$. Then $(\Delta, \preceq)$ is a partial order.
Let $B=\bigcup\limits_{i\in I}B_i$, $f:B\to A^2$ be given by $f(x) = f_i(x)$ if $x\in B_i$. Then $(B,f)\in \Delta$ and is an upper bound for $\Gamma$. Hence, by Zorn's lemma, there exists a maximal element in $\Delta$, say $(M, g)$.
Case 1: $M=A$. Then $A=M\preceq A^2$ (but we already know this).
Case 2: $M\subsetneq A$. Then $g$ must be surjective (can be proved by contradiction with $(M,g)$ being a maximal element), and so $g$ is bijective, so that $A^2\preceq_{g^{-1}} A$.
However, this is impossible, since $M\subsetneq A\implies M\times M \subsetneq A\times A$. But $M = A\times A$, so $M\times M$ cannot be properly contained in $A\times A$. Hence, $A\times A\not\preceq A$.
Can someone please tell me what is my error here?
Your mistake is the conclusion in Case 1. In this case, instead of proving that you obtain the conclusion, you just proved that something trivial holds.
So, if Case 1 happens, you didn't prove what you want.