Proof.
By the definition of \({n \choose k}\text{,}\) we have
\begin{equation*}
{n-1 \choose k-1} = \frac{(n-1)!}{(n-1-(k-1))!(k-1)!} = \frac{(n-1)!}{(n-k)!(k-1)!}
\end{equation*}
and
\begin{equation*}
{n-1 \choose k} = \frac{(n-1)!}{(n-1-k)!k!}\text{.}
\end{equation*}
Thus, starting with the right-hand side of the equation:
\begin{align*}
{n-1 \choose k-1} + {n-1 \choose k} \amp = \frac{(n-1)!}{(n-k)!(k-1)!}+ \frac{(n-1)!}{(n-1-k)!\,k!}\\
\amp = \frac{(n-1)!k}{(n-k)!\,k!} + \frac{(n-1)!(n-k)}{(n-k)!\,k!}\\
\amp = \frac{(n-1)!(k+n-k)}{(n-k)!\,k!}\\
\amp = \frac{n!}{(n-k)!\, k!}\\
\amp = {n \choose k}\text{.}
\end{align*}
The second line (where the common denominator is found) works because \(k(k-1)! = k!\) and \((n-k)(n-k-1)! = (n-k)!\text{.}\)