7. Let $k$ be an integer greater than 1, and $a$ be a root of $x^{2}-k x+1=0$. For any natural number $n$ greater than 10, the units digit of $a^{2^{n}}+a^{-2^{n}}$ is always 7, then the units digit of $k$ is $\qquad$.
Given $a^{2}-k a+1=0 \Rightarrow a+a^{-1}=k \Rightarrow a^{2}+a^{-2}=k^{2}-2$, when the unit digit of $k^{2}$ is always 9, the unit digit of $k^{2}-2$ is always 7, which can be achieved when the unit digit of $k$ is 3 or 7. Note that $n>10$, then the unit digit of $k$ can also be 5. For example, $3 \rightarrow 7 \rightarrow 7, 7 \rightarrow 7 \rightarrow 7, 5 \rightarrow 3 \rightarrow 7$, other unit digits cannot form such a cycle. Therefore, the unit digit of $k$ is 3 or 5 or 7.