Show that the 1st to 99th derivative of $\sin (x^{100})$ evaluated at $0$ is $0$ and the 100th derivative at $0$ is non-zero. To argue that the $0$th to $99$th derivative of $\sin (x^{100})$ is zero, should i use the 2 versions of Taylor series and compare the terms in $\sum_{n=0}^{\infty} (-1)^n \frac{x^{200n+100}}{(2n+1)!}$ and $\sum_{n=0}^{\infty} f^{(j)}(0)/j! \cdot x^j$,
where the first summation does not contain the terms $x,x^2,...,x^{99}$, hence the corresponding coefficient in the 2nd summation $f^{(j)}(0)/j! = 0$, which means $f^{(j)}(0) = 0$
Then the coefficient of $x^{100}$ would be $1$ in the first series and $f^{(100)}(0)/100!$ in the second series, and hence $f^{(100)} = 100!$.
 A: Let $f_k(x) = \sin x^k$ for a fixed $k \in \mathbb Z^+$.  Since $$f_1(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!},$$ it follows that $$f_k(x) = f_1(x^k) = \sum_{n=0}^\infty (-1)^n \frac{(x^k)^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty (-1)^n \frac{x^{k(2n+1)}}{(2n+1)!}.$$  By Taylor's theorem about $0$, $$f_k(x) = \sum_{m=0}^\infty \frac{f^{(m)}(0)}{m!} x^m,$$ thus equating terms gives $$f^{(m)}(0) = \begin{cases} 0, & m \ne k(2n+1) \\ \frac{(-1)^n (k(2n+1))!}{(2n+1)!}, & m = k(2n+1). \end{cases}$$ Thus the smallest value of $m$ for which $f^{(m)}(0) \ne 0$ is when $n = 0$, i.e., $m = k$.
A: How about this: since $x \mapsto x^{100}$ is continuous at zero, we can substitute into
$$ \lim_{y \to 0} \frac{\sin{y}}{y} = 1 $$
to produce
$$ \lim_{x \to 0} \frac{\sin{(x^{100})}}{x^{100}} = 1. $$
Hence
$$ \sin{(x^{100})} = x^{100} + h(x), $$
where $h(x)/x^{100} \to 0$ as $x \to 0$. It follows that the first 99 derivatives of the left-hand side at zero are zero, since the first 99 of the right-hand side are zero, by inspecting successive difference quotients. The hundredth derivative is $100!$, of course, since the hundredth derivative of $x^{100}$ is $100!$ and the hundredth derivative of $h$ is still zero by the only property we know $h$ has.
