Choosing the bilinear form for elliptic equations Evans - First edition - p345 - Problem 6.6.2

The problem statement:

A function $u\in H_0^2(U)$ is a weak solution of this boundary value problem:
  $$\begin{cases}\Delta^2u=f&\text{ in }U\\u=\frac{\partial u}{\partial \nu} =0&\text{ on }\partial U\end{cases}$$
  provided
  $$\int_U \Delta u \Delta v dx= \int_U fv dx$$
  for all $v\in H^2_0(U)$. Given $f\in L^2(U)$, prove that there exists a unique weak solution of the above.


Some notes that give the solution to this problem choose the bilinear form:
$$B[u,v]=\int_U \sum_{i,j=1}^n u_{x_ix_j}v_{x_ix_j} dx,$$
But I can't see how this was chosen.
I thought the idea was to take $Lu=f$ and multiply by $v$, so that we get $L(u)v=fv$, and then integrate over this to define:
$$B[u,v]=\int_U L(u)v dx = \int_U fvdx$$
Where for us $L=\Delta^2$, so that we get:
$$B[u,v]=\int_U \sum_{i,j=1}^n \partial_i\partial_i\partial_j\partial_j(u)vdx =\int_U fv dx$$
Where we can use integration by parts, noting that since $v\in H^2_0$ it vanishes on the boundary, so we get:
$$B[u,v]=-\int \sum_{i,j=1}^n\partial_i\partial_i\partial_j(u)\partial_j(v) dx $$
And we could pass the second $j$ through, but we cannot commute these differential operators, to obtain what they are using?
 A: \begin{align} 
B[u,v]& =-\int_U \sum_{i,j=1}^n\partial_i\partial_i\partial_j(u)\partial_j(v) \mathrm dx =\int_U \sum_{i,j=1}^n -\partial_j\big( \partial_i\partial_i u\partial_j(v) \big) + \partial_i\partial_i u \partial_j \partial_j v \mathrm dx \\
&= \int_U -\nabla \cdot \big( \Delta u \nabla v \big) + \Delta u \Delta v \mathrm dx = \int_{\partial U} \big( \Delta u \nabla v \big) \cdot \boldsymbol{\nu} \mathrm d s + \int_U \Delta u \Delta v \mathrm dx \\
&=  \int_U \Delta u \Delta v \mathrm dx
\end{align}
where the last step stems from the fact that $ \frac{\partial v}{\partial \boldsymbol{\nu}} = \nabla v \cdot  \boldsymbol{\nu} = 0$.
Now you have to do the usual Lax-Milgram business to prove existence & uniqueness.
Continuity:
\begin{align}B[u,v] = & \int_U \Delta u \Delta v \mathrm dx = (\Delta u, \Delta v)_{L^2(U)} \\
\overset{\text{Cauchy-Schwarz}}{\leq} &\Vert \Delta u \Vert_{L^2(U)}  \Vert \Delta v \Vert_{L^2(U)} \leq \Vert u \Vert_{H_0^2(U)}  \Vert v \Vert_{H_0^2(U)} \end{align}
Coercivity:
\begin{align}B[u,v] &= \Vert \Delta u \Vert^2_{L^2(U)} \\
 &= \frac{1}{3} \bigg( \Vert \Delta u \Vert^2_{L^2(U)} + \Vert \nabla \cdot \nabla u \Vert^2_{L^2(U)} + \Vert \nabla \cdot \nabla u \Vert^2_{L^2(U)} \bigg) \\
&\geq \frac{1}{3} \bigg( \Vert \Delta u \Vert^2_{L^2(U)} + C_1 \Vert \nabla u \Vert^2_{L^2(U)} + C_1 \Vert \nabla u \Vert^2_{L^2(U)} \bigg) \\
&\geq \frac{1}{3} \bigg( \Vert \Delta u \Vert^2_{L^2(U)} + C_1 \Vert \nabla u \Vert^2_{L^2(U)} + C_1C_2\Vert u \Vert^2_{L^2(U)} \bigg) \\
&\geq \frac{1}{3} \min \{1, C_1, C_1C_2 \} \Vert u \Vert^2_{H_0^2(U)} \end{align}
where in steps 2,3 the (Poincare-)Friedrichs inequality was used.
