Full binary tree node count

Structural induction. Base case: a single-node tree has $0$ internal nodes and $1$ leaf. $0+1=1$. ✓

Inductive step. Suppose the tree has root with left subtree $L$ and right subtree $R$. Let $n_L,n_R$ be internal node counts and $\ell_L,\ell_R$ leaf counts. By induction, $\ell_L=n_L+1$, $\ell_R=n_R+1$. Total internal nodes: $n=n_L+n_R+1$ (including root). Total leaves: $\ell=\ell_L+\ell_R=n_L+1+n_R+1=n+1$.

Conclusion. A full binary tree with $n$ internal nodes has $n+1$ leaves.