Let $f(x)$ be a polynomial in $F_q[x]$. If we know $\alpha$ is a root, then $\alpha^q$, $\alpha^{q^2}$, ... are roots But why do we assume some $\alpha^{q^r}$ will be equal to $\alpha$. Why does it need to form a full cycle?
Could it not be the case that at some $r$, $\;\alpha^{q^r}$ becomes equal to some $\alpha^{q^j}$, $1 \leq j < r$, and $j$ to $r$ forms a (smaller) cycle?
 A: We in fact can show something a bit stronger:
THM 1. First, let $f$ be an irreducible polynomial in $\mathbb{F}_q[x]$ and denote as $n$ be the degree of $f$. Then let $\alpha$ be a root of $f$. Then $\alpha^{q^n} = \alpha$, and there is no positive integer $m<n$ such that $\alpha^{q^m} = \alpha^{q^j}$ for some nonnegative integer $j<m$.
Proof:. First, note that the field $\mathbb{F}_q(\alpha)$ generated by $\alpha$ has $q^n$ elements, and $(\mathbb{F}(\alpha))^{\times}$ has $q^n-1$ elements. Now, $(q,q^n-1) = 1$, so there is some positive integer $r$ such that $q^r \pmod {(q^n-1)} = 1$ [and in fact this integer $r$ is infact $n$]. So $\alpha^{q^n} = \alpha^{q^n \pmod{q^n-1}} = \alpha$. Thus, in the sequence $S=\{\alpha^{q^i}\}; i = 0,1,2, \ldots$, it follows that $\alpha^{q^r} = \alpha$ for each $r$ that is a multiple of $n$. So if there is some positive integer $m<n$ such that the equation $\alpha^{q^m} = \alpha^{q^j}$ for some nonnegative $j<m$, then the equation $\alpha^{q^j}=\alpha$ must hold. [Indeed, let $j$ be the largest integer less than $m$ such that the equation $\alpha^{q^j} = \alpha^{q^m}$ holds. From this it follows that every element in the sequence $\{\alpha^{q^i}\}; i = j,j+1,j+2, \ldots$ is in the set $\{\alpha^{q^j}, \alpha^{q^{j+1}}, \ldots, \alpha^{q^m}\}$ which does not contain $\alpha$. This contradicts what we observed earlier, that $\alpha = \alpha^{q^r}$ for each $r$ that is a multiple of $n$.]
So now let $m$ be the smallest positive integer such that $\alpha^{q^m} = \alpha^{q^j}$ for some  nonnegative $j<m$. As $\alpha^{q^n}=\alpha=\alpha^{q^0}$, it follows that $m$ must be no larger than $n$. From the above paragraph, $j$ must also be $0$. We now show that $n$ must be $m$. To this end, we first note the following: let $$Q=Q(\alpha,\alpha^{q}, \ldots,\alpha^{q^i},\ldots, \alpha^{q^{m-1}})$$ be any symmetric polynomial in $\alpha^{q^i}; i=0,\ldots, m-1$. Then
$$(Q(\alpha,\alpha^{q}, \ldots,\alpha^{q^i},\ldots,\alpha^{q^{m-2}}, \alpha^{q^{m-1}}))^q$$
$$=Q(\alpha^q,\alpha^{q^2}, \ldots,\alpha^{q^{i+1}},\ldots, \alpha^{q^{m-1}},\alpha^{q^m})$$
$$=Q(\alpha^q,\alpha^{q^2}, \ldots,\alpha^{q^{i+1}},\ldots, \alpha^{q^{m-1}},\alpha^{q^{m-1}},\alpha)$$
$$= Q(\alpha,\alpha^{q}, \ldots,\alpha^{q^i},\ldots, \alpha^{q^{m-1}}),$$
because $Q$ is symmetric. Thus, $Q$ must be in $\mathbb{F}_q$ itself, because all $q$ roots of the polynomial $x^q-x$ are in $\mathbb{F}_q$.
So now let us consider the coefficients of the polynomial $P(x)$ [in $x$] where
$$P(x) = \prod_{i=0}^{m-1} (x-\alpha^{q^i}).$$ Then $P(x)$ is of degree $m$, has $\alpha$ as a root. Also, writing $P(x) = \sum_{i=0}^m c_ix^i$, as each coefficient $c_i$ of $P(x)$ is a symmetric polynomial in $\alpha^{q^i}; i=0,\ldots, m-1$, it follows that $c_i^q=c_i$, and thus $c_i$ is a root of $x^q-x$ and is thus in $\mathbb{F}_q[x]$. So there is a polynomial of degree $m \le n$ that has $\alpha$ as a root; namely, $P(x)$. Thus, as $f$ is an ireducible polynomial in $\mathbb{F}_q[x]$ that has $\alpha$ as a root and the inequality $m \le n$ holds, it follows that $m$ must be $n$, and the result follows. $\surd$
A: Note that $F_{q^t}$ is a field with $q^t$ elements. Thus by Lagrange we get $\alpha^{q^t-1} = 1$. Multiplying both sides with $\alpha$ will give you the desired property.
