Lemma: $\ker(\beta^{n})\subseteq \ker(\beta^{n+1})$. I will take this for granted, unless you want a proof.
Now, write the minimal polynomial of $\alpha$ as
$$
p(t) = (t - \lambda)^rq(t)
$$
where $(t - \lambda) \not \mid q(t)$. Now, since $p(\alpha) = 0$, we have $(\alpha - \lambda I)^r q(\alpha) = 0$, which is to say that $q(\alpha)V \subset \ker(\alpha - \lambda I)^r$.
One direction:
Now, suppose there were an $r'<r$ with $\ker(a -\lambda I)^{r'} = \ker(a - \lambda I)^r$. It would follow that $q(\alpha)V \subset \ker(\alpha - \lambda I)^{r'}$, which would imply that
$$
(\alpha - \lambda I)^{r'} q(\alpha) = 0
$$
which contradicts our definition of a minimal polynomial.
The other direction:
Now, suppose that there were some $R > r$ with $\ker(A - \lambda I)^r \neq \ker(A - \lambda I)^R$. By our lemma, we deduce that $\ker(A - \lambda I)^r \subsetneq \ker(A - \lambda I)^R$. So, there is a vector $v$ such that $(A - \lambda I)^{r'}v = w \neq 0$, but $(A - \lambda I)^{r'+1} v = (A - \lambda I)w= 0$ for some $r' \geq r$.
However, we know that $q(\alpha)(\alpha - \lambda I)^r = 0$, which means that
$$
q(\alpha) w = p(\alpha)v = 0v = 0
$$
That is, $w \in \ker(q(\alpha))$. We also have $w \in \ker(A - \lambda I)$.
That is, $\alpha w = \lambda w$, and $q(\alpha)w = q(\lambda) w = 0$. Thus, we conclude that $q(\lambda) = 0$, which is a contradiction of our definition of $q(t)$.
The conclusion follows.
Note that if we may use Jordan canonical form here, this amounts to a proof by computation.
