I got a little confused with a question about cardinal exponentiation:
Let $\beta$ be an ordinal, and let $K_\alpha$ , $\alpha < \beta$, be infinite cardinals with $K= \sum_{\alpha<\beta}K_\alpha$ and now I have to prove that for every cardinal $\lambda$ $$\lambda^K =\prod_{\alpha<\beta}\lambda^{K_\alpha}$$ Now I know that $k^{\lambda}k^{\mu}=k^{\lambda+\mu}$ for cardinals $k,\lambda,\mu$, so my first attempt was trying induction on $\beta$. However I got stuck when I assumed $\beta$ was a limit ordinal. Am I on the right track with induction or does anyone suggests an alternative way to prove this?
You have $\kappa = \sum_{\alpha < \beta} \kappa_{\alpha}$, so let $X$ be a set of cardinality $\kappa$, and $X_{\alpha}$ of cardinality $\kappa_{\alpha}$. Then $$X = \bigsqcup_{\alpha < \beta} X_{\alpha}$$ is a disjoint union.
Let $Y$ be a set of cardinality $\lambda$.
A function $X \to Y$ is specified precisely by its values on each component $X_{\alpha}$ of $X$. That is, given a function $f: X \to Y$, we may produce a unique member of $$\prod_{\alpha < \beta} \{ \text{functions $X_{\alpha} \to Y$ }\}$$ by $$\langle \text{$f$ restricted to $X_{\alpha}$} \rangle_{\alpha < \beta}$$
This mapping $Y^X \to \prod_{\alpha < \beta} Y^{X_{\alpha}}$ is bijective.