Using the well ordering principal you can prove what you want to prove in the following way:
The principal says that:
For all sets S, S N and S=/= Φ ( S being non empty) S has a least element.
Since the above holds for all non empty subsets of N and since N is non empty and a subset of itself then N has a least member.
Alternatively we can prove the above by induction in the following way:
for n=1 we have .............1>=1 and the theorem is true for n=1
suppose now the theorem true for n=k and k>=1 .And we need to prove that : k+1>= 1.
But k>=1 ====> k+1>= 1+1 >1 +0, since 1>0 and so
................k+1>1 =====> k+1>=1 and the theorem is proved