How to prove that "i & (i - 1) == 0" means "i is the power of 2"

If we know beforehand that i must be something like 10...0, it's not hard to imagine.
But how to mathematically prove that i & (i - 1) == 0 is sufficient to say i is the power of 2, at lease for positive integer?
You can find some explanations of this expression in How to check if a number is a power of 2 But I don't think an explanation is congruent with a mathematical proof.

Technically, the statement is not correct, in case of i is a signed int it should be

i & (i - 1) == 0 if and only if i is a power of 2 or 0

The proof is easy:

If i is power of 2, then it can be represented as 1000...000 (binary) and we have

  1000...000 - i is a power of 2 
  0111...111 - i - 1
  ----------
  0000...000 = 0

If i is 0 then we have

          0 - i = 0
  111...111 -1 (2 complement)
  ---------
          0

If i is not a power of 2 or 0 then it has at least two bits set, let's take leftmost and rightmost of them:

  i = 1?..?10...0
      ^    ^ rightmost 1
      leftmost 1

and we have the leftmost bit ever set:

  1?..?10...0 - i
  1?..?01...1 - i - 1 
  -----------
  1?..?00...0 > 0