*If $k$ is an integer greater than 1, $\alpha$ is a root of $x^{2}-k x+1=0$, for any natural number $n$ greater than 10, the unit digit of $\alpha^{n}+\alpha^{-2^{n}}$ is always 7. Find the unit digit of $k$.*
Let the unit digit of $\alpha^{\alpha^{n}}+\alpha^{-2^{n}}$ be $a_{n}$. Since
$$
\alpha^{2^{n}}+\alpha^{-2^{n}}=\left(\alpha^{n^{n-1}}+\alpha^{-2^{n-1}}\right)^{2}-2 \text{, }
$$

then
$$
a_{n}=a_{n-1}^{2}-2(\bmod 10) .
$$

By Vieta's formulas, $\alpha+\alpha^{-1}=k$, so $a_{1}=\alpha^{2}+\alpha^{-2} \equiv k^{2}-2(\bmod 10)$.
When the unit digit of $k$ is even, by (2), $a_{1}$ is even. By (1), all $a_{n}$ are even. When the unit digit of $k$ is 1 or 9, by (2), $a_{1}=9$. By (1), all $a_{n}=9$. When the unit digit of $k$ is 5, $a_{1}=3$,
$a_{2}=7$, and for $n \geqslant 2$, $a_{n}=7$. When the unit digit of $k$ is 3 or 7, $a_{1}=7$, and all $a_{n}=7$.
Therefore, the answer is that the unit digit of $k$ is $3,5,7$.