Solutions

We prove this by induction. For the base case $n=1$, we have

$$f^{(1)}(x)=-nx^{-n-1}=(-1)^1\frac{(n+1-1)!}{(n-1)!}x^{-n-1}.$$

For the induction step, suppose the equation holds for $k\in\mathbb{N}$. Then,

$$\begin{align*} f^{(k+1)}(x)&=\frac{d}{dx}\left((-1)^k\frac{(n+k-1)!}{(n-1)!}x^{-n-k}\right) \\ &=(-1)^k\frac{(n+k-1)!}{(n-1)!}(-n-k)x^{-n-k-1} \\ &=(-1)^{k+1}\frac{(n+k-1)!}{(n-1)!}(n+k)x^{-n-(k+1)} \\ &=(-1)^{k+1}\frac{(n+k)!}{(n-1)!}x^{-n-(k+1)}. \end{align*}$$