(Milnor's Lemma) Every simple root of $f (t)$ is a super-attractive fixed-point of $N_f (t)$ where $N_f (t) = t - \frac{f (t)}{\dot{f} (t)}$ since a superattractive fixed-point is one such that its multiplier $\lambda_f(t) = \dot{f} (t)= 0$ so that its multiplicity is \begin{equation} m_f (t) = \frac{1}{1 - \lambda_{N_f} (t)} = \frac{1}{1 - 0} = \frac{1}{1} = 1 \end{equation}
John Milnor. Dynamics in One Complex Variable. Annals of Mathematics Studies 160. Princeton University Press, 2nd edition, 2006. Problem 4-g p.54
Proof. Let $\alpha$ be a root $f (\alpha) = 0$ then the multiplier of its Newton map is $$\lambda_{N_f} (\alpha) = \frac{f (\alpha) \ddot{f} (\alpha)}{\dot{f} (\alpha)^2} = 0$$ since $f (\alpha) = 0$ the entire expression $\frac{f (\alpha) \ddot{f} (\alpha)}{\dot{f} (\alpha)^2}$ is equal to 0 since due to the ordering of operations the value of $\dot{f} (t)$ or $\ddot{f} (t)$ is never required to be known in order to know the value of $\lambda_{N_f} (t)$ when $f (t) = 0$.
If any term in the product is $0$ then the entire
product takes the value $0$. The multiplicity is related to the multiplier by
$$m_f (t) = \frac{1}{1 - \lambda_{N_f} (t)} = 1$$ and therefore simple. Since
$$m_f (t) = \frac{1}{1 - \lambda_{N_f} (t)} \forall \lambda_{N_f} (t) \neq
1$$ then it is known that $\lambda_{N_f} (t) = 0$ when $f (t_{}) = 0$
therefore the point $\alpha$ is a superattractive fixed-point corresponding to
a simple zero at $\alpha$.
Since we now know that $m_f = (\alpha)$ and therefore the fero at $f (\alpha) = 0$ is simple, we therefore know that the denominator $\dot{f} (t)$ of the multiplier $\lambda_{N_f} (t)$ cannot vanish so that $\dot{f} (\alpha) \neq 0$ since that would imply that $\alpha$ is not a simple root, which would be a contradiction to the already established fact that $m_f (\alpha) = 1$ when $f (\alpha) = 0$.
If $m=m_f(α)$, then $f$ factorizes as $f(x)=(x-α)^mg(x)$ with $g(α)\ne 0$. Then $$ N_f(x)=x-\frac{(x-α)^mg(x)}{m(x-α)^{m-1}g(x)+(x-α)^mg'(x)}\\ =α+(x-α)\left(\frac{m-1}{m}\right)+O((x-α)^2) $$ so that indeed the stated connection exists, at simple roots the linear coefficient is zero, Newton's method is quadratic. And conversely, if at a root of a sufficiently smooth function the linear coefficient in the expansion of the Newton operator vanishes, then the multiplicity is one, the root is simple.