Trivial normal bundle $NS$ equivalence I am trying to prove the following assertion:

Suppose S is a properly embbeded submanifold of $\mathbb R^n$ of
  codimension $k$. Show that the following are equivalent:
  
  
*
  
*There exists a neighborhood $U$ of $S$ in $\mathbb R ^n$ and a smooth function $\Phi: U \rightarrow \mathbb R^k$ such that $S$ is a regular
  level set of $\Phi$.
  
*The normal bundle $NS$ is a trivial vector bundle
  

This is an exercise of the book Introduction to Smooth Manifolds - John M. Lee and I don't have a clue how to start to prove this.
 A: We are in $\mathbb{R}^n$ all the way, so the computations are more concrete.
To see why $(1) \implies (2)$, we can use the fact that we have a very explicit derivative for $\Phi$:
$$\Phi'_p=\begin{pmatrix} \nabla \Phi_1 \\
\nabla\Phi_2 \\
\cdots \\
\nabla \Phi_k
\end{pmatrix}. $$
Since we are supposing $S$ is a regular level set, we have that those $\nabla \Phi_i$ are all linearly independent along $S$. They are also all normal to $S$, since $\Phi$ is constant there. This gives a global framing of the normal bundle.
To see why $(2) \implies (1)$, you simply use the fact that by assumption there exists a diffeomorphism $\Psi: NS \to S \times \mathbb{R}^k$, and consider $\Phi:=\pi_2 \circ \Psi \circ T,$ where $T: U \to V \subset NS $ is a diffeomorphism of a neighbourhood of $S$ onto a neighbourhood of the zero section on the normal bundle (such diffeomorphism is given by the tubular neighbourhood theorem).
A: $1\Rightarrow 2:$ Since $S$ is a level set of only a single function, that means it has codimension $1.$ Since it is a level set of the function $\Phi,$ that means $d\Phi$ is a differential form which vanishes on all tangent vectors to $S$, and does not vanish for any vectors not tangent to $S$. Using an inner product, we may turn $d\Phi$ into a vector field, nowhere vanishing, and normal, along $S$. Hence the rank one normal bundle is trivial.
$2\Rightarrow 1:$ Take a global section $v$ of $NS$. Define $\Phi(p,t) = \exp(tv_p)$.
