Direct proofs $ 2m < 2^m $ and $ 2^m \not\le m^2 $ for infinite cardinals

Question

I am after the simplest and most direct proofs available that $2\mathfrak{m} < 2^\mathfrak{m} $ and $ 2^\mathfrak{m} \not\le \mathfrak{m}^2 $ for infinite $ \mathfrak{m} $ given GCH but not AC.

This is in connection with proving GCH implies AC in a Kelley-Morse setting, and I am after the tersest most direct proof I can get assuming Foundation and that the theory of ordinals and cardinals have already been defined and developed to some elementary level without AC.

This is a result of Specker, and there are several articles on generalisations and more powerful results. See for example Andrés E. Caicedo which almost gives me what I am looking for.

I hope to avoid a lot of lemmas and theorems around choice and well-orders and get straight to these two results. I have not been able to figure out how to do this easily, although this paper seems close Kanamori and Pincus

In ZF there is more than one "version" of GCH. One is that if $A$ is Tarski-infinite and $A\leq' B\leq' P(A)$ then $A\sim B$ or $A\sim P(A), $where $A\leq'B$ means there is an injection $f:A\to B,$ and $A\sim B$ means there is a bijection $g:A\to B,$ and $P(A)$ is the power-set of $A. $ Another is that if $K$ is an infinite cardinal ordinal then $P(K)\sim K^+.$ — DanielWainfleet, Jan 09 '18 at 15:46
@Daniel: They turn out to be all equivalent, as they all imply the axiom of choice... What's Tarski-infinite, though? — Asaf Karagila, Jan 10 '18 at 06:28
@Daniel: Yes. I can imagine that would be the definition. Let's go one step further. What is "Tarski-finite"? — Asaf Karagila, Jan 10 '18 at 06:38

Direct proofs $ 2m < 2^m $ and $ 2^m \not\le m^2 $ for infinite cardinals

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