I would like to show:
for all $n\in\omega$, $(\aleph_{n+1})^{\aleph_0}\leq\aleph_{n+1}\cdot(\aleph_n)^{\aleph_0}$
Actually I see this problem can be solved using the approach Brian Scott uses here:
For every $n < \omega$, $\aleph_n^{\aleph_0} = \max(\aleph_n,\aleph_0^{\aleph_0})$
Most unfortunately, Brian hasn't been here for over a year (hope he's OK), so perhaps someone would explain one aspect of his proof:
Now consider a function $\varphi:\aleph_0\to\aleph_m$; $\aleph_m$ is an uncountable regular cardinal, so $\sup\{\varphi(k):k\in\aleph_0\}<\aleph_m$, and $\varphi$ actually maps $\aleph_0$ into $\eta$ for some ordinal $\eta<\aleph_m$.
My question is why, since $\aleph_m$ is an uncountable regular cardinal, its cofinality is $\aleph_m$, so why is the $\text{sup}$ less than $\aleph_m$?
Suppose for a contradiction that $\sup \{\varphi(k):k\in\omega\}$ is $\aleph_m$. Then, the sequence of sets $\bigcup_{k\in\omega}\{\varphi(j):j\leq k\}$ is cofinal in $\aleph_m$ and has length $\aleph_0$. This contradicts the regularity of $\aleph_m$ ($m\neq 0$ since $\aleph_m$ is an uncountable cardinal).