Question regarding exponentiation of cardinal numbers

Question

Let $2\leq \kappa\leq \gamma$ where $\gamma$ is an infinite cardinal number and $\kappa$ is any cardinal. Prove that $2^\gamma = \kappa^\gamma$.

One direction is obvious but I'm stuck with the other. i.e. showing $\kappa^\gamma\leq 2^\gamma$.

Any hints or solutions will be appreciated.

Answer 1

First conclude that $\gamma^2=\gamma$. Thus, $2^\gamma=2^{\gamma^2}=(2^\gamma)^\gamma \ge \gamma^\gamma \ge \kappa^\gamma$.

Answer 2

For $0<k\leq l$ with $l$ infinite, show there is an injection $g:l\times k\to l.$

For $f\in^ lk$ let $J(f)=\{g(\langle x,f(x)\rangle): x\in l\}.$ Then $J:^lk\to P(l)$ is injective.

(.....$^lk$ is the set of functions from $l$ to $k,$ and $P(l)$ is the power-set of $l$).