6.9 A calculator has a square key (i.e., pressing this key will make the input number become its square) and a reciprocal key (i.e., pressing this key will make the input number become its reciprocal). If a person starts by inputting a number $x$ not equal to zero, and then alternately presses the square key and the reciprocal key $n$ times each. Assuming the calculator is perfectly accurate, then $y$ should be
(A) $x^{(-2)^{n}}$.
(B) $x^{2 n}$.
(C) $x^{-2 n}$.
(D) $x^{1-2 n}$
(E) $x^{\left[(-1)^{n} 2 n\right]}$.
(35th American High School Mathematics Examination, 1984)
[Solution] Squaring a number can be seen as taking its 2nd power; and finding the reciprocal of a number can be seen as taking its -1st power. Therefore, pressing either of the two keys on the calculator can be regarded as taking the input number to a certain power.
Each press of the reciprocal key changes the sign of the exponent of $x$;
Each press of the square key multiplies the exponent of $x$ by 2.
Thus, pressing the square key and the reciprocal key together is equivalent to multiplying the exponent of $x$ by -2.
Since the initial exponent of $x$ is 1, its final exponent is $(-2)^{n}$.
Therefore, the answer is $(A)$.