Premise. I was given the following definition of beth-numbers:
$$ \beth_\alpha := \begin{cases} \beth_0 = \aleph_0 \\ \beth_{\alpha + 1} = 2^{\beth_\alpha} \\ \beth_{\lambda} = \bigcup_{\alpha < \lambda} \beth_\alpha &\text{$\lambda$ limit} \end{cases}$$
besides, my teacher defined cofinality both as the least cardinality of a cofinal subset of a poset and, (in the case of cardinal numbers), given a cardinal $\kappa$, as the least cardinal $\lambda$ such that there exists a family of cardinals $\{\kappa_i\}_{i \in \lambda}$, $\forall i \in \lambda \; \kappa_i < \kappa$, for which $\kappa = \sum_{i \in \lambda} \kappa_i$. Then he proved (assuming AC) that in the case of infinite cardinals the definitions above are equivalent.
I've came up with some questions I cannot completely find answer to so I'm asking for some help.
I tried to evaluate $\text{cof}(\beth_\lambda)$, where $\lambda$ is a limit ordinal. My claim is that $\text{cof}(\beth_\lambda) = |\lambda|$, and I've been able to prove that $\text{cof}(\beth_\lambda) \leq |\lambda|$, since the following set $\{\beth_\alpha : \alpha < \lambda\}$ is a cofinal subset of $\beth_\lambda$ and it's easy to show that it's cardinality is $|\lambda|$, but I cannot find an easy way to prove the opposite inequality.
It easy to prove by induction that the map $\alpha \mapsto \beth_\alpha$ is non decreasing, but the definition of the beth-numbers gives me the impression that it is possible to prove that the map above is strictly increasing, am I wrong?