The transcendence degree of either $\mathbb{C}$ or $\mathbb{Q}_p$ over $\mathbb{Q}$ is exactly the cardinality of the continuum. Certainly it can't be countable, since any field with countable transcendence degree over a countable field is countable. On the other hand, it has cardinality either transcendence degree is at most that of the continuum. So we're already done assuming CH. But I think the result holds even without CH; can anyone confirm / deny? (Edit: Yes, as the other answer shows.)
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