Cardinality of a vector space versus the cardinality of its basis

Let $V$ an infinite dimensional vector space.

How to show that the cardinality of $V$ is the same of a basis of $V$?

I saw this argument here link in the main answer (MathOverFlow).

You need the additional assumption that $\dim_F V \ge |F|$ (where $F$ is the base field).

Let $V$ be a vector space over a field $F$ and let $B \subseteq V$ be a basis. Suppose $B$ is infinite. There is a surjection $(F \times B)^{<\omega} \to V$ given by $$(( a_i, v_i) : i < n) \mapsto \sum_{i<n} a_iv_i$$ So we have $$|V| \le |(F \times B)^{<\omega}| = \sum_{n<\omega} |F \times B|^n \le \aleph_0 \cdot |F| \cdot |B| = \max \{ \aleph_0, |F|, |B| \}$$ So if $B$ is infinite and $|B| \ge |F|$ then we get $|V| \le |B| = \dim_F V$. (And obviously $|V| \ge \dim_F V$ so we have equality.)