Q:

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

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 $(r-1)!$. 

A:

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.