Example 11. Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)$ ( $n$ is a natural number). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) 1991
(B) $-1991^{1}$
(C) $(-1)^{\star} 1991$
(D) $(-1)^{\star} 1991^{-1}$
Let $a=x+\sqrt{1+x^{2}}, b=x-\sqrt{1+x^{2}}$. Then $a+b=2 x, a b=-1$ or $a=-b^{-1}$. Therefore, $b-b^{-1}=2 x$.
And $2 x=1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}$,
so $b-b^{-1}=1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}$.
Solving, we get $b=1991^{\frac{1}{n}}$ or $b=-1991^{-\frac{1}{n}}$.
Since $b=x-\sqrt{1+x^{2}}<0$, which does not meet the problem's conditions, we discard it.

From $b=-1991^{-\frac{1}{n}}$, we get $b^{n}=(-1)^{n} 1991$. That is, $x-\sqrt{1+x^{2}}=(-1)^{n} 1991$.
Therefore, the answer is (D).