Cardinal exponentiation problem from Halmos' Naive Set Theory 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.        
 A: If $a=b$, there’s nothing to prove, so assume that $a<b$. From the first part you know that $a^c\le b^c$. Since $a$ is finite and greater than $1$, and $b$ is finite, there is a positive integer $n$ such that $a^n\ge b$, and it follows from the first part that $b^c\le (a^n)^c$. If you can show that $a^c=(a^n)^c$, the result will follow from the Schröder-Bernstein theorem. 
To show this, show that $nc=c$ for finite, non-zero $n$ and infinite $c$, and show that in general $(x^y)^z=x^{yz}$. I don’t own a copy of Halmos, but judging from what I can see at Google books, the latter has been stated but not proved. You should go ahead and prove it; there’s a very natural bijection between $\left(X^Y\right)^Z$ and $X^{Y\times Z}$. The former seems not to have been stated yet, but it can be proved by induction on $n$. The only hard step is showing that $2c=c$; after that the induction step just uses that same result. Just after the problem on which you’re working, Halmos proves that $c+c=c$, and there’s a natural bijection between $c+c$ and $2c$, so everything that you need is there.
