First notice that $\alpha$ must be rational, otherwise $\{a_n\}=\{\alpha\lfloor \alpha^n\rfloor\}$ cannot be periodic, so $a_n$ can't either. Now rewrite \begin{align*}a_n = \alpha \lfloor \alpha^n \rfloor- \lfloor \alpha^{n + 1}\rfloor &=(\alpha(\alpha^n-\{\alpha^n\}))-(\alpha^{n+1}-\{\alpha^{n+1}\}) \\ &= \{\alpha^{n+1}\}-\alpha\{\alpha^n\}. \end{align*}From the periodicity, for any $n\ge 1$, we have $$\{\alpha^{n+1+p}\}-\alpha\{\alpha^{n+p}\}=\{\alpha^{n+1}\}-\alpha\{\alpha^n\},$$which is equivalent to $$\{\alpha^{n+1+p}\}-\{\alpha^{n+1}\}=\alpha(\{\alpha^{n+p}\}-\{\alpha^n\}).$$By easy induction we conclude that for any $m,n\ge 1$, $$\{\alpha^{n+m+p}\}-\{\alpha^{n+m}\}=\alpha^m(\{\alpha^{n+p}\}-\{\alpha^n\}).$$Since $\alpha >1$ and the absolute value of left hand side must not exceed 1, taking $m$ sufficiently large will imply that $\{\alpha^{n+p}\}-\{\alpha^n\}=0$ for all $n$. This means $\alpha^n(\alpha^p-1)\in \mathbb{Z}$ for all $n$. However if $\alpha$ is not an integer, let $\alpha=x/y$ and $\alpha^p-1=u/v$ for some positive integers $x,y,u,v$ with $(x,y)=(u,v)=1$ and also $y>1$. Then for any $n\ge 1$, $x^nu$ is divisible by $y^nv$, furthermore $u$ is divisible by $y^n$ since $(x,y)=1$. Take $n$ sufficiently large leads to a contradiction.