Cardinality of k-bijections

Question

Suppose that $k$ is an infinite cardinal, how can I prove that $|\{f:k\rightarrow k :\text{$f$ is a bijection}\} | = 2^{k}$ ?

Answer 1

Choice lets us well-order $k$ and choose its elements' images under $f$ in ascending order, with $k$ unused values available at each point. So there are $k^k$ permutations. But $2^k\le k^k\le (2^k)^k=2^{k^2}=2^k$, completing the proof (note $k^2=k$ follows from choice too).