= Solution
Put $\kappa=\aleph_\omega$. Apply <König theorem for cardinal numbers> to the sequences $\aleph_n<\aleph_{n+1}$:
$$
\kappa=\sum_{n<\omega}\aleph_n
<\prod_{n<\omega}\aleph_{n+1}\leq\kappa^{\aleph_0}.
$$
The sum is $\kappa$ because the countable <union> has <cardinality> at most $\kappa$ and its summands are cofinal in $\kappa$. Thus the power is strictly larger than $\kappa$, a fact that cannot be replaced by the regular-cardinal calculation of the preceding part.
For an upper bound, use $\kappa\leq2^\kappa$ and the <currying law for cardinal exponentiation>:
$$
\kappa^{\aleph_0}\leq(2^\kappa)^{\aleph_0}
=2^{\kappa\cdot\aleph_0}=2^\kappa=\kappa^+,
$$
where the last step is GCH at $\kappa$. There is no cardinal strictly between $\kappa$ and its successor, so
$$
\boxed{\aleph_\omega^{\aleph_0}=\aleph_{\omega+1}.}
$$
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