I want to show that, assuming AC, that if $\kappa$ is a regular cardinal then $\kappa^{<\kappa} = \max\{\kappa, 2^{<\kappa}\}$ where $\kappa^{<\lambda}$ is defined by $$\kappa^{<\lambda} = \sup\{\kappa^{\nu}\mid\nu \in Card , \nu < \lambda\}.$$
If $\kappa= \aleph_0$ then $\aleph_0^{<\aleph_0} = \sup\{\aleph_0^n\mid n \in \omega\} =\aleph_0^{\aleph_0} = 2^{\aleph_0} = \max\{\kappa, 2^{\kappa}\} = \max\{\kappa,2^{<\kappa}\}$.
If $\kappa$ is a successor cardinal, then $\kappa = \aleph_{S(\alpha)} = \aleph_{\alpha + 1}$ and, using Hausdorff's formula, we have $$\aleph_{\alpha+1}^{<\aleph_{\alpha +1}} = \sup\{\aleph_{\alpha+1}^{\aleph_{\beta}} \mid \beta < \alpha + 1\} = \sup\{\max\{\aleph_{\alpha+1},\aleph_{\alpha}^{\aleph_{\beta}}\} \mid\beta < \alpha + 1\} = \max\{ \aleph_{\alpha+1}, \sup\{{\aleph_{\alpha}^{\aleph_{\beta}}\mid\beta < \alpha + 1\}}\} = \max\{\aleph_{\alpha+1},\aleph_{\alpha}^{\aleph_{\alpha}}\}$$ and, since $\aleph_{\alpha}^{\aleph_{\alpha}} = 2^{\aleph_{\alpha}} = 2^{<\kappa}$ this case is proved.
I cannot prove the result for $\kappa$ limit. If $\kappa$ is limit and regular cardinal then it is weakly inaccessible, i.e. $\kappa = \aleph_\kappa$. I'm asking an hint for this part and if the rest of the proof is correct.