$(r-1)^{th}$ derivative of $x^{k+r-1}$

Question

EDIT: added $x^k$ in final answer

I want to find:

\begin{align} \frac{d^{r-1}}{dx^{r-1}}\left(x^{k+r-1}\right) \end{align}

Writing out the first few terms and what I think is the last term we get:

$= (k+r-1)\times(k+r-2)\times(k+r-3)\times\cdots\times(k+1)x^k = \frac{(k+r-1)!}{k!} x^k$

However, my professor told me that the correct answer should be $(r-1)!$.

Answer 1

Let $r=2,\,k=1$.

Then $$x^{k+r-1}=x^2$$ and $$\frac{\mathrm d^{r-1}}{\mathrm dx^{r-1}}\left(x^2\right)=\frac{\mathrm d}{\mathrm dx}\left(x^2\right)=2x.$$

This proves $\left(r-1\right)!$ is not an $(r-1)$-th derivative of $x^{k+r-1}$ by counterexample.

Answer 2

the correct answer is $\frac{d^{r-1}(x^{k+r-1})}{dx^{r-1}} = \frac{(k+r-1)!}{k!} x^k$