Given $\lambda$ regular cardinal, $\left(\kappa^{<\lambda}\right)^{<\lambda}=\kappa^{<\lambda}$?

Note that since $\lambda$ is regular, for any $\mu<\lambda$, $f\colon\mu\to\lambda$ is bounded.

Now think about $g\in\left(\kappa^{<\lambda}\right)^{<\lambda}$ as some $g\colon\mu\to\kappa^{<\lambda}$. Then there is some $\nu<\lambda$ such that $g\colon\mu\to\kappa^\nu$. So we get the wanted result, since clearly $\left(\kappa^{<\lambda}\right)^\mu=\kappa^{<\lambda}$ for any $\mu<\lambda$.

I want to add some details to Asaf's aswer and slightly modify its final argument:

Suppose $\lambda$ weakly inacessible and $\text{cof}(k^{<\lambda})=\lambda$ (the other cases are dealt in the body of the question), then, if we have $g \in \left(k^{<\lambda}\right)^{<\lambda}$ with $g:\mu \longrightarrow k^{<\lambda}$, $g$ must be bounded in $k^{<\lambda}$ (because of its cofinality),
hence $\exists \nu < \lambda$ s.t. $g: \mu \longrightarrow k^\nu$

So we have $$\left(k^{<\lambda}\right)^{<\lambda} = \left|\bigcup_{\mu,\nu<\lambda}\left(\kappa^\nu\right)^\mu\right| = \kappa^{<\lambda} $$