Are there infinite cardinals $\kappa$, $\lambda$ with $\kappa^\lambda = \kappa$? Going through an exam question for revision.
I need to prove the following:

Are there infinite cardinals $\kappa$, $\lambda$ with $\kappa^\lambda = \kappa$?

I really am not sure, though intuition says no. 
I am then asked to state and prove Cantor's Theorem, i.e. there is no surjection from a set to its power set. This is also simple to prove.
The final part of the question asks 

Let $\kappa$ be a cardinal number. Prove $2^\kappa ≠ \aleph_0$

This I am also basically clueless for. Perhaps I can show that for any subset of the naturals, its power set has size either less than or greater than $\aleph_0$, depending on whether or not the subset is finite? But I am not sure if this would even work, let alone constitute a proof.
Any hints anyone can provide would be helpful, since it would help me learn better if I can come across the solutions to these by myself!
Thanks

Yes. Try $\kappa=2^{\aleph_0}$ and $\lambda=\aleph_0$. 
We need to biject sequences of 0-1 sequences with 0-1 sequences. But sequences of sequences are just a 2-dimensional array, and this can be treated wit Cantor's zigzag enumeration.

If $\kappa$ is finite, then $2^\kappa$ is finite. If $\kappa\ge\aleph_0$, then $2^\kappa\ge 2^{\aleph_0}$.
