@m8rsh: You did a bitwise OR instead of a bitwise AND.
0000 0000 0000 0000 0000 0000 0000 0010
AND
0000 0000 0000 0000 0000 0000 0000 0001
Gives
0000 0000 0000 0000 0000 0000 0000 0000

Also, please remember to click the contains spoiler link if you are going to post info about the answer.

BTW, the second explanation answer fails for the reason that both sides will always equal 2n – 1.

An answer without a special case (or at least, a hidden one!) could be:
return !(n^(n-1) < n);

Easy I think. You just have to do a binary AND of the number n with (n – 1). If the result is 0, then n is a power of 2.

For example, 16 is 10000 in binary, 16-1=15 is 01111. And, of course, 10000 AND 01111 is 0 which indicates that 16 is a power of two.

Another example: if n is 12 (not a power of two), it is 1100. 11 is 1011. Anding both gives 1000 which is not 0 hence 12 is not a power of 2.

Hmm. The equal signs didn’t post in my previous comment, guess I don’t know how this editor works. That should read something like:
return n equals equals 2 …

Just my attempt to violate the spirit of the puzzle while still solving it:
return n2 || n4 || n8 || n16 || n32 || n64 …

;)

For any n? What about 2^183842? Mathematically this answer works, but you need arbitrarily large integral types to be able to properly solve this problem.

(in Haskell, not ( n .& (n-1)) /= 0 && n /= 0) would work for all n of type Integer)

(n % 2) ? 0 : 1;

(n & (n ^ (n – 1))) == n;

Your “another way to do this” does not seem to work.

if you were allowed to use external functions…
is_int(ln x/ln 2)