Given a binary tree $t$, prove that $size(t) < 2^{h(t)}$
If a binary tree has height $h$ then it has at most 1 node at depth 0, at most 2 nodes at depth 1, ..., at most $2^{h-1}$ nodes at depth $h-1$ (the maximal depth), and so at most $1+2+\cdots+2^{h-1} = 2^h-1$ nodes in total.

You can also prove this by induction. When $h=1$, the tree consists only of a root, so at most $1=2^h-1$ vertices. Given a tree of height $h$, its two subtrees have height at most $h-1$, and so together contain at most $2(2^{h-1}-1)$ vertices. Together with the root, the tree contains at most $2(2^{h-1}-1)+1=2^h-1$ vertices.