How to prove a complete binary tree of depth N has $2^{N+1} - 1$ nodes?
Hint 1: Draw some binary trees of depth $0, 1, 2$ and $3$. Depth $0$ is only the the root. 

Hint 2: Use Induction on the depth of the tree to derive a proof. 

The base case is depth $n=0$. With depth $0$ we only have the root, that is, $2^{0 + 1} - 1 = 1$ nodes, so the formula is valid for $n=0$.  

The next case is depth $n=0+1=1$, since the tree is full the root must have $2$ leafs. So we have $1 + 2 = 3$ nodes in total. The formula still holds since $2^{1+1} - 1 = 3$. 

Can you generalize this procedure? Does this hold for all $n \geq 0$?