Consider a binary search tree $T$ whose keys are distinct. Show that if the right subtree of a node $x$ in $T$ is empty and $x$ has a successor $y$, then $y$ is the lowest ancestor of $x$ whose left child is also an ancestor of $x$. (Recall that every node is its own ancestor)
The successor is clearly not in the left child, as those elements are smaller. We can then start examining the parents. If $x$ is on the right child of the parent, $x$ is going to be maxim element in the parent. We can continue considering parents, this way, until we reach one, $K$, that has the current tree on the left. We know that $x < K.key$. We also know that $K.right$ will contain bigger elements. So far $K.key$ is the smaller element we've encountered, bigger than $x$.
If we keep going up the tree, we will either find ourselves on the left site of the parent $P$, which means $K.key < P.key$ or the right side, in which case we know that $P.key < x$.