Can zero be in the least element in a proof of mathematical induction?

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The Well-Ordering Principle states that every non-empty set of positive integers has a least element. However, would this also apply to any non-empty set of non-negative integers (including zero)? I don't see why not, but I wanted to check in case I've missed something.

I wanted to prove that for every n (where n = greatest power of 2 in the left binary number in the equation), the following base 2 number equivalences hold. For example:

1 = 0 + 1 (all numbers in base 2).

10 = 01 + 1

100 = 011 + 1

1000 = 0111 + 1

10000 = 01111 + 1

etc...

Since the lowest power of 2 in the left binary number is 0 (see the first equation) the least element in my proof must be n = 0.

I was able to successfully prove it with Induction, assuming that the least element is allowed to be zero.