Q:

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
  $\mu^\lambda=\mu$ ?

A:

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: $\aleph_{\beta+1}^{\aleph_\alpha}=\aleph_\beta^{\aleph_\alpha}\cdot\aleph_{\beta+1}$.

