Prove that $ \delta _{ik} \delta _{kj} = \delta _{ij} $

Question

This is a neat Kronecker-Delta Identity, but I don't know how to go about proving it. Intuitively, I understand that all non-zero contributions to this sum are when $ i = j = k $, but the answer that I saw was that if we say $ \delta _{kj} = A_{k} $, then we can simply write this as $ \delta _{ik} A_{k} = A_{i} = \delta _{ij} $. However, that substitution seems arbitrary - why $ A_{k} $ instead of $ A_{j} $ ?