(n)2-1

2^(N+1)-1..
 
if N=0; it is 2-1=1,1 is the max no of node in the tree
if N=1; it is 4-1=3, 3 is the max no of nodes in the tree
if N=2; it is 8-1=7, 7 is the max

and it goes like that...........

2powN-1

(2 pow n)-1,if root level is one and (2 pow n+1)-1, if root 
level is zero.

(2^(N+1))-1
Suppose level is 2 then total number of nodes will be
1 root
2 left of root and right of root
2 left and right of left of root
2 left and right of right of root
so total nodes are 1+2+2+2=7

by formula (2^(2+1))-1
            8-1=7   

2^(N+1)-1

if the tree is binary tree [i.e two children max]then 
2pow(N) is the answer
if it has 3 max children then
3pow(N)
...............................
if it has n max children then
npow(N)

2^N-1

to be more generic this kind of problem is best solved 
recursivly.

To point out that 2 ^ N AND 3 ^ N are both wrong,
here's a few examples: (the exponet is the amount of levels)
2^0 = 1, correct
2^1 = 2, incorrect, should be 3
2^2 = 4, incorrect, should be 7

And a tree with three children
3^0 = 1, correct
3^1 = 3, incorrect, should be 4
3^2 = 9, incorrect, should be 13

Looking at that I'm sure you can see the pattern.
Let
C = "Number of Possible Children"
N = Levels

N
Σ   C^N
j=0

or in C++ code
int NodeCount(int C, int N)
{
  if (N < 0) return 0
  return NodeCount(C, N-1) + C^N
}

2^N -1