In Jech, Set Theory (1978), Solovay's Theorem is proved (Theorem 81, page 405). In the proof (page 407) we read:
Let $\kappa$ be a compact cardinal. If $\lambda>\kappa$ is an arbitrary cardinal, then we have $$\lambda^{<\kappa}\leq(\lambda^+)^{<\kappa}=\lambda^+$$ In particular, we have $\lambda^{\aleph_0}\leq \lambda^+$ for every $\lambda >\kappa$.
Up to here it is all clear. Now:
By Theorem 23 (or rather by Lemma 8.3), this implies that the singular cardinal hypothesis holds for every $\lambda>\kappa$.
Now can someone help me completing the proof? I recall that Lemma 8.3 says:
Let κ be a singular cardinal, let cf κ > ω, and assume that $λ^{cf\ κ} < κ$ for all $λ < κ$. If ${κ_α : α < cf\ κ}$ is a normal sequence of cardinals such that $\lim\ κ_α = κ$, and if the set $$\{α < cf\ κ : κ_\alpha^{cf κ_α} = κ_α^+\}$$ is stationary in $cf\ κ$, then $κ\ cf\ κ = κ^+$.
For $cf\ \lambda= \aleph_0$ the first quote gives us the result. For $cf\ \lambda>\omega$ we should apply the lemma. The problem is that I don't see how to build the normal sequence.
The set of all ordinals of countable cofinality is stationary in any ordinal of uncountable cofinality.
In combination of Lemma 8.3 this implies that SCH for countable cofinality implies SCH anywhere.