= Solution
Apply <König theorem for cardinal numbers> with $I=\omega$, $\kappa_i=\aleph_i$, and $\lambda_i=\aleph_{i+1}$. Standard <infinite cardinal arithmetic> gives
$$
\sum_{i<\omega}\aleph_i=\aleph_\omega,
$$
whereas every factor in the product is at most $\aleph_\omega$. Hence
$$
\aleph_\omega
<\prod_{i<\omega}\aleph_{i+1}
\leq(\aleph_\omega)^{\aleph_0}
\leq(\aleph_\omega)^{\aleph_\omega}.
$$
Now suppose $2^{\aleph_\omega}=\aleph_\omega$. The exponent laws for infinite cardinals would give
$$
(\aleph_\omega)^{\aleph_\omega}
=(2^{\aleph_\omega})^{\aleph_\omega}
=2^{\aleph_\omega\cdot\aleph_\omega}
=2^{\aleph_\omega}
=\aleph_\omega,
$$
contradicting the strict inequality. Therefore
$$
\boxed{2^{\aleph_\omega}\ne\aleph_\omega.}
$$
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