cardinal number arithmetic regarding exponentiation I am reading a textbook and it is stated that 

we know that $\aleph_0\lt\aleph_0^{\aleph_0}=c$, but that $c^{\aleph_0}=c$. so we might be tempted to conjecture that if $m \gt \aleph_0$, then $m^{\aleph_0}=m$. But this is false. In fact, for every cardinal number $k$, it is possible to find an $m\gt k$ for which $m^{\aleph_0}\gt m$ and also to find a $p\gt k$ for which $p^{\aleph_0}=p$.

I think there is a typo in the book, does the author mean the following statement?

But this is false. In fact, for every cardinal number $k$, it is possible to find an $m\gt k$ for which $m^k\gt m$ and also to find a $p\gt k$ for which $p^k=p$.

Also can you come up with an example where 

$m\gt k$ and $m^k>m$

I know the above statement is true if $k\lt m\lt \aleph_0$ for example $3^2\gt3$. Is there any example where $m\gt k\ge \aleph_0$ for the above statement to be true?
Thanks.
 A: No, the author likely meant the statement as written. It is true, and it makes sense in context.
Let me expand a bit on the author's statement. You have learned that $\aleph_0 < \aleph_0^{\aleph_0}$. That is: there are more sequences of natural numbers than natural numbers. You might find this surprising, or not, depending on your previous intuition about infinity. In a way it makes sense: infinite sequences of something have a lot more degrees of freedom than single elements of that same thing.
Then you learn that $\mathfrak c^{\aleph_0} = \mathfrak c$. So apparently for $\mathfrak c$, this much larger cardinal, it is no longer true that taking infinite sequences of elements of $\mathfrak c$ produces more sets. This might lead you to believe one of two things: (1), maybe $\mathfrak c$ is an unusual exception, and it's the only cardinal for which this is true, or there are only a couple more; or (2), maybe $\mathfrak c$ is just so large that taking infinite sequences doesn't "help" anymore. (For comparison, you might know that for any finite $n > 0$, and any infinite cardinal $\kappa$, it is true that $\kappa^n = \kappa$; maybe the same is true for $n = \aleph_0$ if you take $\kappa$ large enough.)
The author then dispels both these potential misconceptions: no matter how large a cardinal $\kappa$ you take, you can always find two cardinals $\mu, \nu > \kappa$ such that $\mu^{\aleph_0} = \mu$ and $\nu^{\aleph_0} > \nu$. That is, whether or not exponentiation by $\aleph_0$ adds new elements is not just determined by size: as cardinals get larger and larger, you will always find examples of both behaviors.
Edit: You mention in a comment that the author delays the proof. I think this is likely for good reason: the obvious way I would prove the above fact would be via König's theorem, but likely you have not even learned what cofinality is. However, one of the parts is pretty easy to prove: namely that you can find $\mu > \kappa$ with $\mu^{\aleph_0} = \mu$.
Proof: let $\lambda$ be any cardinal greater than $\kappa$, and define $\mu = \lambda^{\aleph_0}$. Then we have that
$$
\mu^{\aleph_0} = \left(\lambda^{\aleph_0}\right)^{\aleph_0} = \lambda^{\aleph_0 \times \aleph_0} = \lambda^{\aleph_0} = \mu.
$$
Here, we have only used the definition of $\mu$, the fact that $\left(\alpha^\beta\right)^\gamma = \alpha^{\beta \times \gamma}$, and that $\aleph_0 \times \aleph_0 = \aleph_0$, both of which are pretty easy to prove if you have not already done so.
