Step 1. We show that there is a positive integer $l$ such that $\alpha^l$ is rational. Say the sequence is of length $k-1$. For any positive integer $n$ the sequence $\{\alpha^{nk}\}$, $\{\alpha^{nk+1}\}$, ..., $\{\alpha^{nk+k-1}\}$ contains two equal elements. Hence, there are infinitely many pairs $i, j$, $0 < i - j < k$, such that $\{\alpha^i\} = \{\alpha^j\}$, i.e. $\alpha^j(\alpha^{i-j} - 1)$ is an integer. Since there are finitely many possible values of $i-j$, at least one of them occurs infinitely often. So we can find $m$ such that $\alpha^j(\alpha^m - 1)$ is an integer for infinitely many $j$. We can divide two such numbers to get $\alpha^l$ is rational for some positive integer $l$.

Step 2. Conclusion. Now if $\alpha^l$ is not an integer, say, is equal to $\frac{a}{b}$ for $b > 1$, $\text{gcd}(a, b) = 1$, then $\{\alpha^{ln}\}$ is an irreducible fraction with denominator $b^n$. This is true for any $n$ so we get infinitely many values, contradiction. So $\alpha^l$ is an integer. If $\alpha$ is irrational, then $\alpha^{nl+1} = \alpha^{nl} \cdot \alpha$ is irrational for any natural $n$ and have distinct fractional parts (Indeed, $\alpha^{il+1} - \alpha^{jl+1} = \alpha(\alpha^{il} - \alpha^{jl})$ cannot be an integer). This contradicts finiteness as well. Thus $\alpha$ is rational, and similar to above we can conclude it is an integer.