Let $\kappa$ be a cardinal number. I want to show that if $\kappa$ is regular and if $2^{<\kappa} = \kappa$ then $\kappa^{<\kappa}= \kappa$. Here is what I got so far: $$ \begin{align}\kappa^{<\kappa} ~&=~ \left| \bigcup\limits_{\alpha < \kappa}\bigcup\limits_{\beta < \kappa}\beta^\alpha \right| \\&\leq~ \sum\sum\limits_{\alpha, \beta < \kappa} |\beta|^{|\alpha|} \\&\leq~ \sum\sum\limits_{\alpha, \beta < \kappa}2^{|\beta|^{|\alpha|}} \\&=~ \sum\sum\limits_{\alpha, \beta < \kappa}2^{|\beta| \cdot |\alpha|} \\&=~ \sum\limits_{\gamma < \kappa}2^\gamma \\&=~ \kappa \cdot 2^{< \kappa} \\&\hspace{-3pt}\overset{\text{Hyp}}{=} \kappa \cdot \kappa \\&=~ \kappa. \end{align}$$ But I feel that the first line is wrong...
The following seems right $$\kappa^{< \kappa} ~=~ \left| \bigcup\limits_{\alpha < \kappa}\kappa^\alpha \right|$$ but I got doubts about $$\kappa^\alpha ~=~ \bigcup\limits_{\beta < \kappa}\beta^\alpha$$ since $\kappa$ is a regular cardinal.
Any clue?
If $\kappa$ is regular, and $\alpha<\kappa$ then every function $f\colon\alpha\to\kappa$ is bounded, so it is in fact a function into some $\beta<\kappa$. So the equality holds.
Assuming that $\kappa$ is regular does in fact finishes the proof. If $\kappa$ is singular the implication is false. For example assuming $\sf GCH$ we have that: $$2^{<\aleph_\omega}=\sup\{2^{\aleph_n}\mid n<\omega\}=\sup\{\aleph_{n+1}\mid n<\omega\}=\aleph_\omega<\aleph_{\omega+1}=\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{<\aleph_\omega}.$$