# Combinatorial identity $\sum_{j=i}^n \begin{pmatrix} j-1 \\ i-1 \end{pmatrix} = \begin{pmatrix} n \\ i \end{pmatrix}$ [duplicate]

prove $$\sum_{j=i}^n \begin{pmatrix} j-1 \\ i-1 \end{pmatrix} = \begin{pmatrix} n \\ i \end{pmatrix}$$ What I am doing now is to time $$i$$ on both side and try to argue out of this: $$\sum_{j=i}^n i \begin{pmatrix} j-1 \\ i-1 \end{pmatrix} = i\begin{pmatrix} n \\ i \end{pmatrix}$$