cardinal numbers proof Suppose $a,b$ are cardinals where $a$ is finite and $b$ is infinite. I want to prove that $b^a=b$. The book gives a hint saying to use repeated multiplication of cardinals to do it. 
I have proved the result using induction and the fact that $b^{a+1}=b^ab^1$ but I am just noticing that we have not proved this yet, in fact its the next question. So is there some way to prove my question without this property?
So I am trying to find the cardinal number of the set of all functions from $A$ to $B$, where $b=|B|, a=|A|.$ 

You need two facts for this proof, one of which is fairly easy to prove by hand.


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*There is a bijection between ${}^AB\times B$ and ${}^{A\cup\{a\}}B$, where $a\notin A$. From this we can conclude that $b^a\cdot b=b^{a+1}$.
This is really just writing down the simplest, most obvious bijection and showing that it is a bijection, which is not very hard.

*If $X$ is an infinite set, then $X\times X$ and $X$ have the same cardinality. Namely, $a=a^2$ for every infinite $a$.
This is a bit trickier, since this statement is in fact equivalent to the axiom of choice, so we can't quite prove it out of the blue in this sort of generality; although we can prove it for some sets, like $\Bbb N$ or $\Bbb R$ for example.
