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I am trying to prove that for every ordinal $\alpha$, there is a cardinal number greater then $\alpha$.

Proof: $\alpha$ is a (well ordered) set. Take the set $P(\alpha)$. Assume that we know that for each set $A$, if $|A| = k$, then $|P(A)| = 2^{k}$, and that we also know that $k<2^k$ for each cardinal $k$. Can we conclude that $|P(\alpha)|$ is a cardinal greater then $\alpha$ and finish the proof?

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Yes. We can. You can also use Hartogs construction, which shows that given a set $X$, there is an ordinal which is not equipotent to any subset of $X$. If $X=\alpha$, then there is an ordinal $\beta$ such that $|\alpha|<|\beta|$, taking the least such $\beta$ would suffice.

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