Prove that: $\aleph_0 \cdot \frak{c} = \frak{c} \cdot \frak{c}$ I've been fiddling with this enough.
Found an answer here but didn't quite understand it.

How do I prove that:
$$\aleph_0 \cdot \frak{c} \leq \frak{c} \cdot \frak{c}$$
$$\frak{c} \cdot \frak{c} \leq \aleph_0 \cdot \frak{c}$$

The first inequality I've proven already, the second one I just cannot get it(obvious is it may be).

The only thing I'm allowed to use is the definition of cardinality, the cantor-bernstein-schroder theorem and the fact that:
$$ k \leq p \wedge m \leq n \to k\cdot m \leq p \cdot n $$
(which is helpful through the definition of cardinality)

Any insights are welcomed!
$\mathfrak{c}\times\mathfrak{c} = 2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0} = 1\times 2^{\aleph_0} \leq \aleph_0\times 2^{\aleph_0}$.

For the equality $2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0}$, note that a pair of functions from $\aleph_0$ to $\{0,1\}$ is equivalent to a function from the disjoint union of two copies of $\aleph_0$ to $\{0,1\}$.

Since $\aleph_0+\aleph_0$ is the cardinality of the disjoint union of two copies of $\aleph_0=\omega$, and $2^{\kappa}$ is the cardinality of the set of all functions from $\kappa$ to $\{0,1\}$, the above argument seems to meet your requirements.
$\mathfrak{c}\times\mathfrak{c} =\text {max}\{\mathfrak{c},\mathfrak{c}\}
=\mathfrak{c}=\text{max}\{\mathfrak{c},\aleph_0\}=\mathfrak{c}\times\aleph_0  $