We need to show \begin{eqnarray*} \binom{n}{k} + \binom{n}{k-1} & = & \binom{n+1}{k} \end{eqnarray*}Let us begin by writing the left-hand side as $$ \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}$$ Getting a common denominator and simplifying, we have \begin{eqnarray*} \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-k+1)!} & = & \frac{(n-k+1)n!}{(n-k+1)k!(n-k)!}+\frac{kn!}{k(k-1)!(n-k+1)!} \\ & = & \frac{(n-k+1)n!+kn!}{k!(n-k+1)!} \\ & = & \frac{(n+1)n!}{k!((n+1)-k)!} \\ & = & \frac{(n+1)!}{k!((n+1)-k)!} \\ & = & \binom{n+1}{k} \end{eqnarray*}