Example 8 Try to find
$$
\begin{aligned}
p= & (1-1993)\left(1-1993^{2}\right) \cdots\left(1-1993^{1993}\right)+1993\left(1-1993^{2}\right)\left(1-1993^{3}\right) \cdots\left(1-1993^{1993}\right)+ \\
& 1993^{2}\left(1-1993^{3}\right) \cdots\left(1-1993^{1993}\right)+1993^{3}\left(1-1993^{4}\right) \cdots\left(1-1993^{1993}\right)+\cdots+ \\
& 1993^{1992}\left(1-1993^{1993}\right)+1993^{1093} .
\end{aligned}
$$
Analysis and Solution Generalization, the $(k+1)$-th term of $p$ has the following form:
$$
p_{k, n}(x)=x^{k}\left(1-x^{k+1}\right)\left(1-x^{k+2}\right) \cdots\left(1-x^{n}\right) \text {. }
$$

Let $\phi_{n}(x)=\sum_{k=0}^{n} p_{k \cdot n}(x)$, then $p=\phi_{1993}$ (1993).
If we can find $\phi_{n}(x)$, the problem can be solved. It is not difficult to see:
(1) When $k<n$, $p_{k, n}(x)=\left(1-x^{n}\right) p_{k, n-1}(x)$;
(2) When $k=n$, $p_{n, n}(x)=x^{n}$.

Thus, there exists a recursive relationship: $\phi_{n}(x)=\left(1-x^{n}\right) \phi_{n-1}(x)+x^{n}$,
Subtracting 1 from both sides, we get $\phi_{n}(x)-1=\left(1-x^{n}\right)\left(\phi_{n-1}(x)-1\right)$,
Iterating $n-1$ times, we obtain $\phi_{n}(x)-1=\left(1-x^{n}\right)\left(1-x^{n-1}\right) \cdots\left(1-x^{2}\right)\left[\phi_{1}(x)-1\right]$.
Since
$$
\phi_{1}(x)=p_{0.1}(x)+p_{1.1}(x)=(1-x)+x=1,
$$

Therefore, $\phi_{n}(x)=1$, and thus we find $p=\phi_{1993}(1993)=1$.