How can I show $n^{\aleph_0} = \mathfrak c$, if $n$ is finite and at least $2$? Also how can I show $\aleph_0 ^{\aleph_0} = \mathfrak c?$
I know there is a theorem thats says $2^{\aleph_0} = \mathfrak c$, from this I said $\mathfrak c = 2^{\aleph_0} \leq n^{\aleph_0}$ because $n$ can be $2$ or $>2$. So $2^{\aleph_0}\leq n^{\aleph_0}$. But how can $n^{\aleph_0}= \mathfrak c$? Isnt $\mathfrak c \leq n^{\aleph_0}$.
Note: This a not a homework question. Its a practice problem. Getting help is allowed.
Remember that $n^{\aleph_0}$ is the cardinality of all the functions from a set of cardinality $\aleph_0$ into a set of cardinality $n$.
For every $n$ consider the set $\{0,\ldots,n-1\}$ and take $\Bbb N$ as a set of cardinality $\aleph_0$. So $n^{\aleph_0}$ is the cardinality of $\{0,\ldots,n-1\}^\Bbb N$.
Obviously, if $2\leq n$ then $\{0,1\}^\Bbb N\subseteq\{0,\ldots,n-1\}^\Bbb N\subseteq\Bbb{N^N}$. Therefore $2^{\aleph_0}\leq n^{\aleph_0}\leq\aleph_0^{\aleph_0}$.
To show equality we first note that $\left(A^B\right)^C\sim A^{B\times C}$, and therefore $\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}$. Since we know that $\aleph_0\cdot\aleph_0=\aleph_0$ it follows that it is equal $2^{\aleph_0}$. Now we have, $$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0}.$$