What is the transcendence degree of the field of meromorphic functions over $\mathbb{C}$?
By a cardinality argument (meromorphic functions are determined by their image under a countable dense subset and $\mathbb{C}^{\mathbb{Q}}$ has the same cardinality as $\mathbb{C}$), I know a transcendence basis has cardinality at most $|\mathbb{C}|$ but I am unsure if it is exactly this size.
Let $\Lambda=\{\lambda_i\}$ be a transcendence basis for $\mathbb{C}$ over $\mathbb{Q}$, and WLOG assume $|\lambda_i|>2$ for all $i$. $\Lambda$ is uncountable, hence (I believe) we can partition $\Lambda$ into uncountably many countable disjoint subsets, say $\{\lambda_{n\alpha}\}_{n=1}^\infty$, where $\alpha\in I$, an uncountable index set.
Now let $p_\alpha(z)=\sum\limits_{n=0}^\infty \lambda_{n\alpha}^{-n!}z^n$. I claim the $p_{\alpha}(z)$ are algebraically independent over $\mathbb{C}$.
For if there were a polynomial relation between $p_1,\dots,p_k$, say $q(p_1,\dots,p_n)=0$, then $q\in\mathbb{C}[x_1,\dots,x_k]$ is really a polynomial in some finitely generated extension of $\mathbb{Q}$ determined by its coefficients, say $q\in\mathbb{Q}(\{\alpha_j\}_{j=1}^m)[x_1,\dots,x_k]$. Only finitely many of the $\lambda_{in}$ will become algebraically dependent over $\mathbb{Q}(\{\alpha_j\}_{j=1}^m)$ hence for some $N$ the collection $\{\lambda_{in}\}_{i=1,\dots,k;\ n>N}$ will be algebraically independent over $\mathbb{Q}(\{\alpha_j\}_{j=1}^m)$. But then $q(p_1,\dots,p_k)=0$ will give algebraic relations between these $\lambda_{in}$ for high enough powers, for a contradiction.
If the above proof is correct, this would give rise to uncountably many algebraically independent meromorphic functions, hence the transcendence degree will uncountable.