Solutions

We prove this with mathematical induction. For the base case, we have

$$S_n(x)=\frac{n!}{(n-0)!}x^{n-0}=x^n.$$

Now suppose the equation holds for $k$. Then,

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