Is the tr.deg of Q_p over Q 1? and what about C over Q?

In both cases the transcendence degree is the cardinality of the continuum. CH is not needed. This is a corollary of the following result: let $K$ be any infinite field, and let $L/K$ be any extension. Then $\# L = \operatorname{max} (\# K, \operatorname{trdeg}_K L)$. To prove this, in turn it suffices to establish the following two results (each of which is straightforward): 1) If $K$ is infinite and $L/K$ is algebraic, then $\# L = \# K$. 2) If $K$ is any infinite field, $T = \{t_i\}_{i \in I}$ is an arbitrary set of indeterminates and $K(T)$ is a purely transcendental function field in the indeterminates $T$, then $ \# K(T) \leq \# T + \# K$. 


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, 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.) 

