Equalities of cardinal numbers

Question

I need prove that:

$2^{\aleph_{0}}=n^{\aleph_{0}}=\aleph_{0}^{\aleph_{0}}=c^{\aleph_{0}}=c$, for $n\geq2$. Where $c$ is the continuum.

I know that $2^{\aleph_{0}}\leq n^{\aleph_{0}}\leq\aleph_{0}^{\aleph_{0}}\leq c^{\aleph_{0}}$ and $2^{\aleph_{0}}=c$ then if $c^{\aleph_{0}}\leq c$ I'll can show what I want. Anybody know how I can do that?

Thanks!

Answer 1

HINT: $(\kappa^\lambda)^\mu=\kappa^{\lambda\cdot\mu}$ for any three cardinals, $\kappa,\lambda,\mu$.