Prove there are 2^2^N functions with N binary bit inputs and one bit output.

Proof by picture:
For 1 bit input, 2^1 = 2 possible input combinations
input| output       2^2^1 = 2^2 = 4 possible functions 
     | f0 f1 f2 f3  
   0 |  0  0  1  1
   1 |  0  1  0  1

For 2 bit input, 2^2 = 4 possible input combinations
input| output       2^2^2 = 2^4 = 16 possible functions
     | f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 fA fB fC fD fE fF
 0 0 |  0  0  0  0  0  0  0  0  1  1  1  1  1  1  1  1
 0 1 |  0  0  0  0  1  1  1  1  0  0  0  0  1  1  1  1
 1 0 |  0  0  1  1  0  0  1  1  0  0  1  1  0  0  1  1
 1 1 |  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1

For 3 bit input, 2^3 = 8 possible input combinations
input | output      2^2^3 = 2^8 = 256 possible functions
      | f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 fA fB fC fD fE fF ... f255
0 0 0 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...   1
0 0 1 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...   1
0 1 0 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...   1
0 1 1 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...   1
1 0 0 |  0  0  0  0  0  0  0  0  1  1  1  1  1  1  1  1 ...   1
1 0 1 |  0  0  0  0  1  1  1  1  0  0  0  0  1  1  1  1 ...   1
1 1 0 |  0  0  1  1  0  0  1  1  0  0  1  1  0  0  1  1 ...   1
1 1 1 |  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1 ...   1

The next would be 4 bit input, thus 2^4 = 16 possible input combinations,
thus  2^2^4 = 2^16 = 65,536 possible functions

The next would be 5 bit input, thus 2^5 = 32 possible input combinations,
thus  2^2^5 = 2^32 = 4,294,967,296 possible functions

QED There are 2^2^N possible functions with N binary bit inputs
and one binary bit output.

These are called "truth tables" and any function can be 
implemented with less than 2^(N-1),  N input "and" gates plus
one "or" or "nor" gate (Use the "nor" gate if more than half
the outputs are '1') A maximum of 2^((N+1)/2)+1 "and" gates.