The Stacks project

The penultimate sentence reads

... becaue the cardinality of a finite product of copies of $F$...

I think this should be

... because the cardinality of a countable product of copies of $F$...,

since the set of nonconstant polynomials with coefficients in $F$ is a subset of $F^{(\mathbb{N})}$

Good catch! See change here.

I think this is still not quite correct, as a countable product of countables is not countable (already a countable product of finite sets isn't countable), so in particular if we take $F=\mathbb{Q}$, we see that a countable product of copies of $F$ has bigger cadinality than that of $F$ (which is equal to $\aleph_0$). Hence this is not a valid argument.

However, we are only interested in a subset of $F^{\mathbb{N}}$, which can be obtained as a countable union (all possible degrees $n$) of finite products (all possible $n+1$ coefficients of the polynomial) of $F$, and this is indeed bounded by $\max{(\aleph_0,|F|)}$.