I am very stuck on how to prove the opposite of what I was given.
Ok so I have two equations that involve Aleph numbers
$\aleph_{1}^{\aleph_{0}} = 2^{\aleph_{0}}$
and
$\aleph_{0}^{\aleph_{1}} = 2^{\aleph_{1}}$
The first one seems easy because by the Continuum Hypothesis
$2^{\aleph_{0}} = \aleph_{1}$
so I just substituted
$\aleph_{1}^{\aleph_{0}} \leq (2^{\aleph_{0}} )^{\aleph_{0}} = 2^{\aleph_{0} \cdot \aleph_{0}} = 2^{\aleph_{0}}$
But for
$\aleph_{0}^{\aleph_{1}} = 2^{\aleph_{1}}$
I am not sure... like I could try and use the Continuum Hypothesis again such that
$2^{\aleph_{0}} = \aleph_{1}$
becomes $\aleph_{0}^{2^{\aleph_{0}}}$ but this is so messy and backwards.
How can I prove that $\aleph_{0}^{\aleph_{1}} = 2^{\aleph_{1}}$ is valid?
CH is a red herring. And a proof using it isn't an outright proof, so you should resist it regardless.
EDIT: Per the comments below, the key point here is that basic cardinal arithmetic is boring: for all infinite $\kappa,\lambda$ we have $$\kappa+\lambda=\kappa\cdot\lambda=\max\{\kappa,\lambda\},$$ and this (or rather, the multiplication part) is the rule which lets us solve the problems above. It's only when we come to exponentiation that interesting things happen.
Your proof of $(1)$ doesn't actually use CH! Regardless of whether CH is true, we know $2^{\aleph_0}\ge\aleph_1$. So the calculation $$(\aleph_1)^{\aleph_0}\le(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$ doesn't need CH. The converse $2^{\aleph_0}\le(\aleph_1)^{\aleph_0}$ is trivial, and so you're done.
For $(2)$, the same basic idea is going to work. This time, use the fact that $2^{\aleph_0}\ge\aleph_0$. This gives us $$(\aleph_0)^{\aleph_1}\le (2^{\aleph_0})^{\aleph_1}.$$ Do you see how to proceed to get $(\aleph_0)^{\aleph_1}\le 2^{\aleph_1}$ (the converse being trivial as before)?