The Singular Cardinal Hypothesis holds above a compact cardinal

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.

• Why??? WHY??? WHY would you read that version of Set Theory, and not new one? The book had gone through two revisions! Oct 23 '17 at 15:13
• Also, since I don't have that version of the book, I don't know if it's just a typo, or if the mistake is you misquoting from the book. But it is impossible that $\lambda<\kappa$ and that $(\lambda^+)^{<\kappa}=\lambda^+$. Specifically because $\lambda^+<\kappa$, so $\lambda^+<(\lambda^+)^{\lambda^+}<\lambda^{<\kappa}$. Oct 23 '17 at 15:16
• Fixed, sorry. Numbering apart, the old is the same as the new in this part. In the new version, it is page 372-373. Oct 23 '17 at 15:20