Let $\kappa > \omega$ be some limit ordinal, and let $L$ be the set of limit ordinals less than $\kappa$. Since $(L,<)$ is well-ordered, it must be isomorphic to some ordinal $\lambda$ via a map $f:L \rightarrow \lambda$.
It is clear that we must have $f(x) \leq x$ for any $x \in L$; however, is it possible that $f$ has a fixed point, i.e. $f(x) = x$ for some $x$? Or do we have $f(x) < x$ for any $x$?
Certainly any regular cardinal greater than $\aleph_0$ will be a fixed point of $f$. If it weren't we would have written it as the union of less than that many ordinals.
The smallest example is $\omega^\omega$. $f(\omega)=0, f(\omega\cdot 2)=1$ and so on, so $f(\omega^2)=\omega.$ Similarly $f(\omega^n)=\omega^{n-1}$. As $f(\omega^\omega)$ must be greater than all the $f(\omega^n)$ it must be $\omega^\omega$.