Connection between finiteness of dimension of a vector space and existence of a polynomial, such that P(T) is a zero map.

Let $V$ be a vector space over a field $F$. Consider the following assertions:

$(I) V$ is finite dimensional.

$(II)$ For every linear transformation $T : V → V ,$ there exists a nonzero polynomial $p(x) ∈ F[x]$ such that $p(T) : V → V$ is the zero map.

Is it true that both statements are equivalent in the sense that one imply the other$?$

Seems true to me. But no idea how do we construct such polynomial, which maps every element of basis of $V$ to $0$. Any hint would be appriciable.

For $(I) \Rightarrow (II)$ you could use that Hom$(V) = \{T: V \to V: f \text{ is a homomorphism of vector spaces}\}$ is also finite dimensional (it has dimension $\dim(V)^2$). That means that for every $T$ the $\dim(V)^2 +1$ elements $$1, T, T^2, T^3, \dots, T^{n^2}$$ are linearly dependent in Hom$(V)$ and therefore there is a linear combination $$a_0 + a_1 T + a_2 T^2 + a_{n^2}T^{n^2}=0$$ where not all $a_i$ are equal to zero. That yields, that $$p(x):= \sum_{i=0}^{n^2} a_ix^i$$ fulfills $p(T) \equiv 0$.

If on the other hand $V$ has infinite dimension, we can fix an inifite set $\{e_i: i \in \mathbb N\}$ of linearly independent vectors (which exists by Zorn's lemma - even in the infinite dimensional case) and define $T:V \to V$ by $T(e_i) = e_{i+1}$.

Then we have that any expression of the form $$a_0 + a_1T + a_2 T^2 + \dots + a_m T^m =: p(T)$$ (with not all $a_i$ equal to zero) will never be the zero-map. Why? because $$p(T)(e_1) = a_0 e_1 + a_1 e_2 + a_2e_3 + \dots + a_m e_{m+1} \ne 0$$ which follows immediately from the fact that the $\{e_i: i \in I\}$ are linearly independent. That proves $(II) \to (I)$.