Proving $|\alpha^{<\omega}| < \kappa$ without Axiom of Choice Let $\kappa$ be a cardinal. Fix an ordinal $\alpha < \kappa$. Let $\alpha^{<\omega}$ denote the set of all finite sequences of ordinals less than $\alpha$. Can we show that $|\alpha^{<\omega}| < \kappa$ without invoking the Axiom of Choice?
I know it is, for instance, independent without $\mathsf{AC}$ that $\omega_1$ is a countable union of countable ordinals, but I'm unsure of variants of this statement (such as the one I posed).
 A: From the well ordering of $\alpha$ we can recursively define an injection of $\alpha^n$ into $\alpha$, and this allows you to define an injection from $\alpha^{<\omega}$ into $\alpha\times\omega$, which is equipotent with $\alpha$.
More explicitly, we know that there is a function $f_2\colon\alpha^2\to\alpha$ which is a bijection: first choose a function $e\colon\alpha\to|\alpha|$, then using the Gödel pairing function we have that $|\alpha|\times|\alpha|=|\alpha|$ and applying $e^{-1}$ gives us this function.
Now, by recursion, if $f_n\colon\alpha\to\alpha^n$ is a bijection, $f_{n+1}\colon\alpha^{n+1}\to\alpha$ is defined by $$f_{n+1}(\xi_0,\dots,\xi_n)=f_2(f_n(\xi_0,\dots,\xi_{n-1}),\xi_n).$$
Now we have a sequence $\langle f_n\mid n<\omega\rangle$ of functions, where $f_1$ is the identity, and we can ignore, for a moment $f_0$, since it only adds a single point to a well-orderable infinite set. And now we can easily map $\vec\xi$ of length $n$ to the pair $\langle f_n(\vec\xi),n\rangle\in\alpha\times\omega$, and finally we have that $\alpha\times\omega$ injects into $\alpha$ since it is a subset of $\alpha\times\alpha$, and we already covered that part.

Alternatively, note that the set is the same in $L$, since finite sequences of ordinals are absolute, and that in $L$ the axiom of choice holds, so $\alpha$ and $\alpha^{<\omega}$ are equipotent in $L$, and therefore also in $V$.
