$\{ \lambda \in Card | \lambda^k > \lambda\}$ is a proper class

Question

Suppose $k$ is an infinite cardinal. I want to prove that the classes $$ A = \{\lambda \in Card | \lambda^k = \lambda\}$$ and $$ B = \{\lambda \in Card | \lambda^k > \lambda\}$$ are proper. For the first class I thought that I can use fixed points of the function $F : Ord \to Ord$ where $F(\alpha) = |\alpha|^k$; since $F$ is continuous and increasing $$\forall \alpha \exists \beta>\alpha(F(\beta)=\beta)$$ Suppose A is a set (of cardinals), then $\bigcup A = \gamma$ is a cardinal and $\exists \beta>\gamma(F(\beta)=\beta)$. It's a contradiction because, since $\beta = |\beta|^k$, $\beta$ is a cardinal that belongs to $A$ and it's bigger than $\bigcup A$. Is it a correct proof? For the other class I have no idea. Some hints?

Answer 1

You can combine two facts that you might know by now:

1. For every infinite cardinal $\mu$, $\mu^{\operatorname{cf}(\mu)}>\mu$.
2. For every regular cardinal $\kappa$, there is a proper class of cardinals with cofinality $\kappa$.

Specifically, this means that if $\operatorname{cf}(\lambda)\leq\kappa$, then $\lambda^\kappa\geq\lambda^{\operatorname{cf}(\lambda)}>\lambda$.