Suppose $\kappa$ and $\lambda$ are infinite cardinals and that $\lambda$ is regular. Kunen states somewhere that this means that we have $\left(\kappa^{<\lambda}\right)^{<\lambda}=\kappa^{<\lambda}$ I want to prove this. The inequality $\geq$ is clear, so let's focus on the other one. We can also assume $\lambda$ is a limit cardinal. I'm going to mix arithmetic and functional notation a bit, because I find it a bit easier to think this way. So, we're looking at $\left(\kappa^{<\lambda}\right)^{<\lambda}=\bigcup_{\mu<\lambda}\left(\mu\to\left(\bigcup_{\nu<\lambda}\left(\nu\to\kappa\right)\right)\right)=(*)$ Since $\lambda$ is regular, for a fixed $\mu$ in the first union, the second union can be shortened to a $\bigcup_{\nu<\zeta(\mu)}$, for some cardinal $\mu\leq\zeta(\mu)<\lambda$. We can now continue the above equality with $(*)=\bigcup_{\mu<\lambda}\left(\mu\to\sum_{\nu<\zeta(\mu)}\kappa^\nu\right)\leq \bigcup_{\mu<\lambda}\left(\mu\to\zeta(\mu)\cdot\kappa^{\zeta(\mu)}\right)=\sum_{\mu<\lambda}\zeta(\mu)^\mu\cdot\kappa^{\zeta(\mu)}$
I don't know how to continue from this point. The expression on the right looks a bit like what I want to get, but I feel like my estimates were a bit too rough. I'm also worried that $\zeta(\mu)$ is unbounded in $\lambda$, but I'm sure something of this sort must come into play, because I don't see any other way to use the fact that $\lambda$ is regular.