> [!NOTE] Theorem > For all $r\in \mathbb{R},$ $k\in \mathbb{Z}$ $k{r \choose k}=r {r-1 \choose k-1}$where ${r\choose k}$ is a [[Binomial Coefficient|binomial coefficient]]. **Proof**: If $k=0$ then $k{r \choose k}= r {r-1 \choose k-1} =0$since $k-1=-1<0$ so ${r-1 \choose k-1}=0$ by definition. Otherwise, we have $\begin{aligned} k\binom{r}{k}&=k\frac{r\left(r-1\right)\left(r-2\right)\cdots\left(r-k+1\right)}{k\left(k-1\right)\left(k-2\right)\cdots1} \\ &=\frac{r\left(r-1\right)\left(r-2\right)\cdots\left(r-k+1\right)}{\left(k-1\right)\left(k-2\right)\cdots1} \\ &=r\frac{(r-1)(r-2)\cdots((r-1)-(k-1)+1)}{(k-1)(k-2)\cdots1} \\ &=r\binom{r-1}{k-1}. \end{aligned}$