Example 6. If $k$ is an integer, the quadratic equation $(k-1) x^{2}-p x+k=0$ has two positive integer roots, find $k^{k 0}\left(p^{p}+k^{k}\right)$. (84 Beijing Mathematics Competition Question)
Let the two roots be $x_{1}, x_{2}$, then $x_{1}, x_{2}$ are both positive integers. By Vieta's formulas, $x_{1} x_{2}=\frac{k}{k-1}$ is also a positive integer. Since $k$ is a positive integer, if $k-1 \neq 1$, then $k-1$ and $k$ are coprime. In this case, $\frac{k}{k-1}$ cannot be an integer, so $k-1=1$, i.e., $k=2$ is the only way for $\frac{k}{k-1}$ to be an integer. At this time, $x_{1} x_{2}=2=1 \times 2$, then $p=3$. Therefore, $k^{k p}\left(p^{p}+k^{k}\right)=2^{2} \times 3\left(3^{3}+2^{2}\right)=1984$.