Specker's Theorem ($\sf{ZF}$) states that for cardinals $\mathfrak{m}\geq5$, $2^{\mathfrak{m}}\not\leq\mathfrak{m}^{2}$, and thus $n\cdot\mathfrak{m}<2^{\mathfrak{m}}$ for all finite $n$.
I tried to generalize this: For all infinite $\mathfrak{m}$ and finite $n$, $2^{\mathfrak{m}}\not\leq\mathfrak{m}^{n}$, and $n\cdot\mathfrak{m}<2^{\mathfrak{m}}$. I am trying to follow the proof from Andrés E. Caicedo's excellent answer here. I have re-worded it to try and make sure I understand it myself.
Let $|S|=\mathfrak{m}$ and assume for the sake of contradiction that there is an injection $f:\mathcal{P}(S)\rightarrow S^{n}$. We will show that every ordinal injects into $S$, contradicting the existence of the Hartogs number of $S$. To begin, we define an injection from $\omega$ into $S$.
Note first that there is some finite $k$ for which $2^{k}>k^{n}$. Since $S$ is infinite, let $S_{k}=\{s_{0},\dots,s_{k-1}\}$ be a set of $k$ elements from $S$. Note also that $\mathcal{P}(S_{k})\subseteq\mathcal{P}(S)$, but since $2^{k}>k^{n}$ and $f$ is injective, the image of $\mathcal{P}(S_{k})$ under $f$ cannot be a subset of $S_{k}^{n}$. Thus, there is some subset of $S_{k}$ which maps to some $n$-tuple not in $S_{k}^{n}$. Since $k$ is finite, well-ordering $\mathcal{P}(S_{k})$ (lexicographically), we may choose the least subset $A_{k}$ of $S_{k}$ for which $f(A_{k})\not\in S_{k}^{n}$. Then if $f(A_{k})=(x_{0},\dots,x_{n-1})$, and since there is a unique well-ordering corresponding to any $n$-tuple (listing the elements from left to right), we let the least $x_{i}$ for which $x_{i}\not\in S_{k}$ be $s_{k}$, and we have thus defined $S_{k+1}$. Continuing this process inductively yields an injection from $\omega$ into $S$.
My first question is about the choice of $A_{k}$ and $s_{k}$. I want to confirm if my reasoning as far as why $\sf{AC}$ is not being used is correct: Since we chose a definable well-ordering of $\mathcal{P}(S_{k})$, we are not making infinitely many different choices for each $S_{k}$. I think a more formal way of putting it is that there is some formula $\phi(S,T)$ which describes a well-ordering $T$ of the power set of any finite $S$. In particular, we do not need to choose one well-ordering for each $\mathcal{P}(S_{k})$ (which would require $\sf{AC}_{\omega}$ at least). Similarly, the well-ordering for an $n$-tuple $(a_{0},\dots,a_{n-1})$ is defined "in the same way for all $n$-tuples", so choosing the least $a_{i}$ satisfying some $\phi(x)$ is not a use of $\sf{AC}$. Is this correct?
Lastly, I'm not sure how to prove $n\cdot\mathfrak{m}<2^{\mathfrak{m}}$. Obviously we have $2^{\mathfrak{m}}\not\leq n\cdot\mathfrak{m}$, else $2^{\mathfrak{m}}\leq n\cdot\mathfrak{m}\leq\mathfrak{m}^{2}$, a contradiction. But I am not sure how to find an injection from $n\cdot\mathfrak{m}$ to $2^{\mathfrak{m}}$.
What I am trying to do is map each ordered pair $(i,m)$ with $0\leq i\leq n-1$ and $m\in\mathfrak{m}$ (abusing notation a bit) to an $i+1$ element subset of $\mathfrak{m}$. There's an obvious injection from ordered pairs $(0,m)$ to the singletons, and it seems very clear to me that there are at least as many $k$-element subsets as there are $1$-element subsets for any positive integer $k$. If I can prove this then composing injections gives the result. But strangely a proof of this "very clear" statement escapes me. How might I create such an injection, even if there is a different way? (If you have another way, please share, but I'm also trying to see if my original argument works.)
