$11 \cdot 14$ Let $S=(x-1)^{4}+4(x-1)^{3}+6(x-1)^{2}+4(x-1)+1$, then $S$ equals
(A) $(x-2)^{4}$.
(B) $(x-1)^{4}$.
(C) $x^{4}$.
(D) $(x+1)^{4}$.
(E) $x^{4}+1$.
(12th American High School Mathematics Examination, 1961)
[Solution] Consider the binomial expansion
$$
(a+1)^{4}=a^{4}+4 a^{3}+6 a^{2}+4 a+1,
$$

Let $a=x-1$, then $S=x^{4}$. Therefore, the answer is $(C)$.