Proving $2^m \not\leq m^2$ for infinite cardinal $m$ without AC How can I prove that $2^m \not\leq m^2$ for infinite cardinal $m$ without using Axiom of Choice? Andrés E. Caicedo's answer in  Question about Generalized Continuum Hypothesis dealt with  this problem, but I want to know the exact method of proving(specifically, the method called 'Specker's original argument' in the answer of linked article). Edited: I want to know if there is a simpler way than 'Specker's original argument'.
 A: You can find the proof in Jech "The Axiom of Choice" as Lemma 11.10. The proof is due to Specker.
The idea is to show that if there was such an injection, then there is an injection from $\aleph(m)$ into $m$, where $\aleph(m)$ is the Hartogs' number of $m$, and that is by definition impossible.
Suppose that $f\colon\mathcal P(m)\to m^2$ is an injection, and fix $\{x_0,x_1,x_2,x_3,x_4\}$ to be some arbitrary four elements of $m$ ($m$ is infinite, but we really just need it to have $5$ or more elements). Now proceeding by recursion, first dealing with the finite index, $n\geq 5$:
Let $C_n$ denote the set $\{x_i\mid i<n\}$, then there is some $U\subseteq C_n$ such that $f(U)\notin C_n\times C_n$. This is because $n^2<2^n$ for finite integers where $n\geq 5$. Because we are given an order on $C_n$ (we are keeping track of the enumeration because this is a recursive process), there is a definable well-ordering of $\mathcal P(C_n)$, so let $U$ be the first set as above. Now let $f(U)=(x,y)$, since $f(U)\notin C_n\times C_n$ either $x$ or $y$ are not in $C_n$. If $x\notin C_n$, let $x_n=x$, otherwise let $x_n=y$.
This leads us to $C_\omega$, which is a countably infinite set. For every $\alpha<\aleph(m)$, let $f_\alpha$ be the canonical bijection from $\alpha$ to $\alpha\times\alpha$. Suppose that $C_\alpha=\{x_\xi\mid\xi<\alpha\}$ was defined, we may assume that $f_\alpha\colon\alpha\to C_\alpha\times C_\alpha$ by applying the function to the indices (which again, due to the recursion process, are being tracked through the construction).
Define $g(\xi)=f^{-1}(f_\alpha(\xi))$, namely decode an ordered pair from $\xi$, and take the preimage of that ordered pair under $f$. We can now diagonalise against $g$, to explicitly define $U\subseteq C_\alpha$ such that $f(U)=(x,y)\notin C_\alpha\times C_\alpha$, and define $x_\alpha$ to be $x$ or $y$ as above.
And this process may continue up to $\aleph(m)$ which defines an injection from the Hartogs' number of $m$ into $m$, which is impossible, of course.
