If $n < \aleph^*(m)$, then $n < 2^m$.

Question

Without $AC$

Let $\aleph^*(m)$ be the least aleph that $\not\leq^* m$. I need a help or hint that if $n < \aleph^*(m)$, then $n < 2^m$.

$a \leq^* b$ means we can define a surjective map from $b$ onto $a$.

Answer 1

HINT: If there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $\mathcal P(A)$. This leaves the case where $n=0$, but that's easy.

Answer 2

Is my proof true? I suppose that $n = 2^m$. Then $m < n$ and so $m <^* n$. From $m <^* n$ and $n \leq^* m$, $m <^* m$, a contradiction.