2008 Calculus #4:
An alternate, slightly cleaner solution (with the same answer) has been
proposed:
Rewrite sin^n(x) = (x - x^3/3! + x^5/5! + ... + (-1)^k x^(2k + 1)/(2k + 1)!)^n.
If we expand the polynomial on the right, note that the term with the smallest
degree is x^n and that any term with a degree greater than n will be equal to
zero after differentiation n times and substituting x = 0. Thus, the nth
derivative of sin^n(x) at x = 0 is equivalent to the nth derivative of x^n at
x = 0, which is n!.
This solution was proposed by Ray Tung on 11 November 2012.