Let $f \in End(V)$; then $f$ is diagonalizable iff its minimal polynomial is "free from squares", as in, all of its terms are all raised to the first power and (edit:) all irreducible factors are linear. I've seen similar questions on this matter bun none of them actually answer my question... I think I've figured out the first implication (which I will write down now), but I fell I'm nowhere near to figure out the second; so any help or suggestion would be truly appreciated!
Here's my take of the first implication:
As previously said let $f \in End(V)$ be diagonalizable of (distinct) eigenvectors $\lambda_1,...,\lambda_k$; then we know that $$(x-\lambda_1)...(x-\lambda_k) \quad \text{divides} \quad q_f(x)$$ the minimal polynomial of $f$.
Showing that $(f-\lambda_1 I)...(f-\lambda_k I)(v)=0 \quad \forall v \in V$ would imply that in the previous statement "divides" could be replaced with "equals"; now we know that $(f-\lambda_1 I)...(f-\lambda_k I)=0 \iff$ it's $0$ on a base for $V$; since $f$ is diagonalizable there exist a base of eigenvectors $v_i$ of eigenvalue $\lambda_i$; hence we have: $$(f-\lambda_1 I)...(f-\lambda_k I)(v_i)=0 \quad \forall v_i$$ since the terms $(f-\lambda_i I)$ commute with each other.
I believe I managed to prove the implication myself as follows:
Let $q_f$ be the minimal polynomial of $f$ as in: $$q_f(t)=(t-\lambda_1)\cdots(t-\lambda_k)$$ We want to prove that $f\in End(V)$ is diagonalizable, so that $V$ is direct sum of eigenspaces. Now let all eigenvalues $\lambda_1, ..., \lambda_k \in \Bbb{K}$ a field. We know that $V$ is direct sum of generalized eigenspaces $V^{\lambda_i}$ and we want to prove that each one of them is actually an eigenspace $V_{\lambda_i}$. So we want to show that if $w \in V^{\lambda_i}$ then $w \in ker(f-\lambda_i I)$. Now, we know that
$$0=q_f(f)=(f-\lambda_1 I)\cdots(f-\lambda_k I)$$
so
$$\begin{align*} 0&=q_f(f)(w)\\ &=(f-\lambda_1 I)\cdots(f-\lambda_k I) (w)\\ &=(f-\lambda_1 I)\cdots(f-\lambda_{i-1} I)(f-\lambda_{i+1} I)\cdots(f-\lambda_k I)(f-\lambda_i I) (w) \end{align*}$$
Since generalized eigenspaces are $f$-invariant, they are $(f-\lambda_i I)$-invariant too, so we can consider:
$$(f-\lambda_1 I)|_{V^{\lambda_i}}\cdots(f-\lambda_{i-1} I)|_{V^{\lambda_i}}(f-\lambda_{i+1} I)|_{V^{\lambda_i}}(f-\lambda_k I)|_{V^{\lambda_i}}(f-\lambda_i I)|_{V^{\lambda_i}} (w)$$
But now we know that $(f-\lambda_j I)|_{V^{\lambda_i}}$ is invertible when $j \neq i$, so from
$$(f-\lambda_1 I)|_{V^{\lambda_i}}\cdots(f-\lambda_{i-1} I)|_{V^{\lambda_i}}(f-\lambda_{i+1} I)|_{V^{\lambda_i}}(f-\lambda_k I)|_{V^{\lambda_i}}(f-\lambda_i I)|_{V^{\lambda_i}} (w)=0$$
we get that $(f-\lambda_i I)|_{V^{\lambda_i}} (w)=0$ which impies $w\in \ker(f-\lambda_i I)$ as needed.