Measurable $\rightarrow$ Mahlo

On page 135 of Jech there is a step that I do not understand in the proof of "measurable implies Mahlo". We have already proved that "measurable implies strong limit". Jech writes

"As $\kappa$ is a strong limit, the set of all strong limit cardinals $\alpha < \kappa$ is closed unbounded...."

Surely every strong limit cardinal does not have a club set of strong limit cardinals below it (take the least strong limit cardinal, for example).

Am I missing something?

thank you

Generally, the set of strong limit cardinals is always closed, because the limit of strong limit cardinals is a strong limit cardinals. It's just bounded in most cases, but since $\kappa$ is regular, we get the unboundedness as well, to see this:
Let $\lambda<\kappa$, take $\lambda_0=\lambda;\ \lambda_{n+1}=2^{\lambda_n};\ \lambda_\omega=\sup\{\lambda_n\mid n\in\omega\}$. Since $\kappa$ is strong limit, we have that $\lambda_n<\kappa$ for all $n$, and because it is [uncountable] regular, $\lambda_\omega<\kappa$ as well.
It is not hard to see that $\lambda_\omega$ is a strong limit cardinal, and therefore the set of strong limit cardinals is unbounded below $\kappa$, so it has to be a club.