Compute the following two cardinals: $\mathfrak{c}^{\mathfrak{c}^\mathfrak{c}}$ and $n^{\aleph _0} \aleph_0 ^n$ I'm really not sure how to go about this, for the first cardinal we essentially end up with $\mathfrak{c}^{2^\mathfrak{c}}$ but I have no idea where to go from there or even if that's the right step.
For the second one however I know that $\aleph_0 ^n=\aleph _0$, but I'm not sure about the other half. If I can prove that has to be = to $\aleph_0$ as well would that mean the proof is done as we'd end up with $\aleph_0 \aleph_0$, which we know is $\aleph_0$?

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*Since $(A^B)^C\cong A^{B\times C}$, we have that $$\mathfrak c^{\mathfrak c^{\mathfrak c}}=\left(2^{\aleph_0}\right)^{\left(2^{\aleph_0}\right)^{2^{\aleph_0}}}=2^{\left(\aleph_0\cdot2^{\left(\aleph_0\cdot 2^{\aleph_0}\right)}\right)}=2^{\aleph_0\cdot2^{2^{\aleph_0}}}=2^{2^{2^{\aleph_0}}}$$


*Your assertion on $\aleph_0^n$ is essentially correct (though you forgot to mention that $\aleph_0^0=1$). However for $2\le n<\aleph_0$ we have that $n^{\aleph_0}=2^{\aleph_0}>\aleph_0$. Therefore,
$$n^{\aleph_0}\cdot\aleph_0^n=\begin{cases}0&\text{if }n=0\\ \aleph_0&\text{if }n=1\\ 2^{\aleph_0}&\text{if }n\ge2\end{cases}$$
