Finding an algebraic proof for $r{n \choose r} = n{n-1 \choose r-1}$ I can't seem figure this proof out.
How are both sides equal.
$$r{n \choose r} = n{n-1 \choose r-1}$$

Here are a couple of approaches in addition to the approach used by Dror and TMM.
Generating Functions:
$$
\frac{\mathrm{d}}{\mathrm{d}x}(1+x)^n=\sum_{r=1}^n r\binom{n}{r}x^{r-1}
$$
$$
\begin{align}
n(1+x)^{n-1}
&=\sum_{r=0}^{n-1}n\binom{n-1}{r}x^r\\
&=\sum_{r=1}^nn\binom{n-1}{r-1}x^{r-1}\\
\end{align}
$$
Compare the coefficients of $x^{r-1}$

Induction: Suppose it is true for row $n-1$ in Pascal's Triangle, then
$$
\begin{align}
r\binom{n}{r}
&=r\left[\binom{n-1}{r}+\color{#C00000}{\binom{n-1}{r-1}}\right]\\
&=r\binom{n-1}{r}+\color{#C00000}{(r-1)\binom{n-1}{r-1}+\binom{n-1}{r-1}}\\
&=\color{#00A000}{(n-1)\binom{n-2}{r-1}+(n-1)\binom{n-2}{r-2}}+\binom{n-1}{r-1}\\
&=\color{#00A000}{(n-1)\binom{n-1}{r-1}}+\binom{n-1}{r-1}\\
&=n\binom{n-1}{r-1}
\end{align}
$$
