# Regular cardinal for a fixpoint of the exponential

Let $$\lambda$$ be a fixed regular cardinal. I would need (for a proof) to find a regular cardinal $$\mu>\lambda$$ such that $$\mu^\lambda=\mu$$. What I can understand is that the mapping $$\mu\mapsto \mu^\lambda$$ has fixpoints. Indeed, $$(2^\lambda)^\lambda=2^{\lambda^2}=2^\lambda$$. But I believe that, for example, the fact that $$2^{\aleph_0}$$ is regular is undecidable in ZFC and $$\aleph_0$$ is regular. So this fixpoint is probably not a good candidate.

Does it exist a regular cardinal $$\mu>\lambda$$ such that $$\mu^\lambda=\mu$$ ?

## 1 Answer

You are correct that $$2^\lambda$$ could be singular. But $$\mu=(2^\lambda)^+$$ is regular. And let's see how it works out:

$$\mu^\lambda=((2^\lambda)^+)^\lambda=(2^\lambda)^\lambda\cdot(2^\lambda)^+=(2^\lambda)^+=\mu.$$

Of course, the second equality is due to Hausdorff's formula for exponentiation: $$\aleph_{\beta+1}^{\aleph_\alpha}=\aleph_\beta^{\aleph_\alpha}\cdot\aleph_{\beta+1}$$.