Example 1 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)(n$ is an integer). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) 1991
(B) -1991
(C) $(-1)^{n} 1991$
(D) $(-1)^{n} 1991^{-1}$
(1991, National Junior High School Mathematics League)
Solution: From the given, let
$$
1991^{\frac{1}{n}}=x+t, -1991^{-\frac{1}{n}}=x-t \text {. }
$$

Then $-1=x^{2}-t^{2}$, which means $t^{2}=1+x^{2}$.
Also, $x+t>x-t$, so $t>0$.
Therefore, $t=\sqrt{1+x^{2}}$.
$$
\begin{array}{l}
\therefore\left(x-\sqrt{1+x^{2}}\right)^{n}=(x-t)^{n} \\
\quad=\left(-1991^{-\frac{1}{n}}\right)^{n}=(-1)^{n} 1991^{-1} .
\end{array}
$$

Hence, the correct choice is (D).