[jay_shark]

Another proof for 1 .

Each face must have at least 3 edges since 1 and 2 edges are degenerate cases . Now we show that for a polyhedron with n faces , there cannot be a face with more than n edges . For if there were , then the polyhedron would have more than n faces . This is easy to check since for each edge there are 2 faces . So 2*(n+1) - (n+1) = n+1

So we have n faces and they may take on the values
{3,4,5,6,...,n} . This means that there exists two faces with the same number of edges .