Prove the next cardinal property: $\kappa>1 \implies \lambda \leq \kappa^{\lambda}$

Question

Let $\kappa>1$ and $\lambda$ be cardinals. Prove that $\lambda \leq \kappa^{\lambda}$.

Answer 1

Suppose you have sets $A,B$ with respective cardinalities $\kappa,\lambda.$ By definition, $\kappa^\lambda$ is the cardinality of $A^B,$ which is the set of functions $B\to A.$ By assumption, $A$ has at least two elements--fix any two, say $y,z.$ Now, given any $x\in B,$ define $f_x:B\to A$ by $$f_x(t)=\begin{cases}y & t=x\\z &t\neq x.\end{cases}$$

Now, what can you say about the map $B\to A^B$ given by $x\mapsto f_x$?

Answer 2

Ok, thank you Git Gud:

$\lambda \leq |P(\lambda)|= 2^{\lambda} \leq \kappa^{\lambda}$, since $2\leq\kappa$

isn't it?