I am stuck at a particular point in a proof. I would like to say ${\aleph _0}^n \leq {\aleph _0}^{\aleph _0}$ which equals the continuum but I have no proposition that allows me to state the inequality even though it seems trivial. How can I show this inequality?
$n$ is just the cardinal of an $n$ element set
In general $\kappa^2 = \kappa$ for any infinite cardinal $\kappa$ (more generally, though provable from that fact, is that for $\kappa,\lambda$ infinite cardinals, $\kappa+\lambda = \kappa \cdot \lambda = \max\{\kappa,\lambda\}$).
Thus, $\aleph_0^n = \aleph_0$.
For an easier proof, recall the Cantor-Schroeder-Bernstein Theorem that if $|A| \leq |B|$ and $|B| \leq |A|$, then $|A|=|B|$. The map $(m_1,\ldots,m_{n}) \mapsto 2^{m_1}\cdot 3^{m_2} \cdots p_{n}^{m_n}$ is an injection, where $p_n$ is the $n$-th prime number (starting at $p_1 = 2$).
For the inequality you're hoping for, however, you can use more basic facts: if $\kappa$ is a nonzero cardinal and $\lambda \leq \mu$, then $\kappa^\lambda \leq \kappa^\mu$. Recalling that $\kappa^\lambda$ is the cardinality of the set of functions from $\lambda$ to $\kappa$, and likewise with $\kappa^\mu$, just note that any map from $\lambda$ to $\kappa$ extends to a map from $\mu$ to $\kappa$ (say, send everything in $\mu \setminus \lambda$ to $0$), and this is an injection.