In chapter 24 of Halmos' Naive Set Theory the following problem is posed as an exercise (page 96):
Prove that if $a, b$ and $c$ are cardinal numbers such that ${a}\le{b}$, then $a^c\le{b^c}$. Prove that if $a$ and $b$ are finite, greater than $1,$ and if $c$ is infinite, then $a^c=b^c$.
My problem lies in proving the second proposition. The identity ${a}\cdot{a}=a$ for infinite cardinal numbers is not covered yet at that point of the text so that the direct proof ${2}\le{a}\le{2^c}=>{2^c}\le{a^c}\le{2^{{c}\cdot{c}}=2^c}$ is not available. Doing a bit of research, wikipedia points to the use of the axiom of choice on its cardinal number article. I let A, B, and C be sets such that card(A)=a, card(B)=b, card(C)=c, and $a\le{b}$. I then tried to apply Zorn's lemma to the set of functions $f$ such that: ${domf}\subset{B^C}$, ${ranf}\subset{A^C}$, and $f$ is one-to-one; which is partially ordered by inclusion. This yields a maximal element say $\phi$. That's where I am stuck.