Suppose $F$ is an infinite set (that is $\#F\geq\#\mathbb N$). Various sources I have consulted claim that $$\# F=\# (F\times\mathbb N)$$ without proof (# denotes cardinality). I guess that this is so trivial that it should be immediate, yet I can't see it.
Obviously, $\# (F\times\mathbb N)\geq\#F$, because the function $(f,n)\mapsto f$ from $F\times\mathbb N$ to $F$ is clearly a surjection, but I can't prove the other direction, that $\#(F\times\mathbb N)\leq\#F$ (only in the particular case when $F$ is countable, i.e., $\#F=\#\mathbb N$). Once I could show this, the desired result would follow from the Schröder–Bernstein theorem.
Any help would be appreciated.
You can't really prove this "outright", because the axiom of choice is necessary for establishing this result.
HINT: Assuming the axiom of choice, we have that $\#F=\#(F\times F)$, and that $F$ is infinite is the same as saying as $\#F\geq\#\Bbb N$.