$V_\omega$ is countable Is there an easy way to prove this? I found a book that suggests the injection $h:V_\omega\to\omega$ defined by
$$h(\{x_1,x_2,\dots,x_n\})=2^{h(x_1)}+2^{h(x_2)}+\cdots+2^{h(x_n)},$$
but I hit some snags while proving that the function is well-defined. Are there any other clever ways to prove this fact? (Note: I am making a formal proof, so I need to be able to justify my steps.)
Edit: Let me see if I can make my difficulty more clear. Before I can start the proof by induction, I need a precise statement of the induction hypothesis, and this is what I am trying to hammer out. The hypothesis:

For each $n\in\omega$, there exists a unique bijection $h_n:V_n\to|V_n|$ such that $$h_n(x)=\sum_{m<|V_{n-1}|}\operatorname{if}(h_{n-1}^{-1}(m)\in x,2^m,0).$$
Moreover, the $h_n$ as defined satisfies $h_n\restriction |V_{n-1}|=h_{n-1}$.

Does this hypothesis contain enough to prove the claim? The reason for the odd way of stating the sum is that it avoids the issue in the linked question (we need to show that addition is a well-defined operation over sets), but as a consequence I have to prove that
the $h_n$ is a bijection and not just an injection, and I think I also need the restriction part in order to prove that the union $h=\bigcup_{n\in\omega}h_n$ is the required bijection.

If you just need to prove that $V_\omega$ is countable, and if you're allowed to use the axiom of choice, then I think you're working too hard.  Prove, by induction on natural numbers $n$ that $V_n$ is finite (because the power set of a finite set is finite).  Then $V_\omega=\bigcup_{n\in\omega}V_n$ is the union of countably many finite sets, hence countable.
