Solutions

Notice that

$$\lim_{h\to 0^-}\frac{f(h)}{h}=\lim_{h\to 0^-}\frac{0}{h}=0=\lim_{h\to 0^+}h^{n-1}=\lim_{h\to 0^+}\frac{f(h)}{h}=f'(0).$$

Hence,

$$f'(x)=\begin{cases} nx^{n-1}, & x\ge 0,\\ 0, & \text{otherwise}. \end{cases}$$

and by Exercise 27 we have

$$f^{(n-1)}(x)=\begin{cases} n!x, & x\ge 0,\\ 0, & \text{otherwise}. \end{cases}$$

But

$$\lim_{h\to 0^-}\frac{f^{(n)}(h)}{h}=\lim_{h\to 0^-}\frac{0}{h}=0\ne n!=\lim_{h\to 0^+}\frac{n!h}{h}=\lim_{h\to 0^+}\frac{f^{(n)}(h)}{h}.$$

Thus, $f^{(n)}(0)$ does not exist.