How can I show that if the second fundamental form of a surface is identically equal to zero, then the surface is a plane? This is my question:
Let P be a plane considered as a surface in 3-space. Show that its second fundamental
form is zero. Conversely, show that if the second fundamental form of a surface is identically zero then the surface is a plane.
So far I have that the general equation of the plane is
$$ax+by+cz+d=0.$$
Then
$$M(u,v)=(u,v,\alpha u + \beta v + \gamma),$$
where $a,b,c,d,\alpha,\beta, \gamma$ are all constants and $c\not=0$.
Then $M_u=(1,0,\alpha)$ and $M_v=(0,1,\beta)$. I know that $\vec{n} =\frac{M_u \times M_v}{|M_u \times M_v|}$ which gives
$$ \vec{n}=\frac{(-\alpha,-\beta, 1)}{\sqrt{\alpha^2+\beta^2+1}}.$$
Hence, the unit norm is a constant.
$$ M_{uu}=M_{uv}=M_{vv}=0$$
Hence $l=m=n=0$, so the second fundamental form is
$$ ldu^2+mdudv+ndv^2=0.$$
Now I need to show the converse, but I don't know how. Please help :)

Proposition 4.2 If the second fundamental form of a surface vanishes, it is part of
a plane.
Proof: 
If the second fundamental form vanishes,
$0 = r_u \cdot n_u =  r_v \cdot n_u = r_u \cdot n_v = r_v \cdot n_v$
so that
$n_u = n_v = 0$
since $n_u$, $n_v$ are orthogonal to $n$ and hence linear combinations of $r_u$, $r_v$. 
Thus $n$ is
constant.
This means
$(r \cdot n)_u = r_u \cdot n = 0$,
$(r \cdot n)_v = r_v \cdot n = 0$
and so
$r \cdot n$ = const
which is the equation of a plane. 
