Evans's proof of the Leibniz's formula for the weak derivatives in Sobolev spaces In PDE Evans 2nd edition, pages 261-263, there is a theorem and its proof which concerns the four properties of weak derivatives. Unfortunately, I do not understand the fourth property, which I will type here accordingly. Note that, according to Evans' notation in his textbook, 


*

*$W^{k,p}(U)$ denotes a Sobolev space that consists of all locally summable functions $u : U \rightarrow \mathbb{R}$ such that for each multiindex $\alpha$ with $|\alpha| \le k$, $D^\alpha u$ exists in the weak sense and belongs to $L^p(U)$.

*$C_c^\infty(U)$ denotes the space of infinitely differentiable functions $\phi : U \rightarrow \mathbb{R}$, with compact support in $U$, and the function $\phi$ is called a test function.



Theorem 1 (Properties of weak derivatives). Assume $u,v, \in W^{k,p}(U), |\alpha| \le k$. Then
$\quad$ (iv) If $\zeta \in C_c^\infty(U)$, then $\zeta u \in W^{k,p}(U)$ and $$D^\alpha (\zeta u)=\sum_{\beta \le \alpha} {\alpha \choose \beta}  D^\beta \zeta D^{\alpha - \beta} u \qquad \textit{(Leibniz' formula)} \tag{7}$$ 
  $\quad$ where ${\alpha \choose \beta} = \frac{\alpha!}{\beta!(\alpha-\beta)!}$.

Again, I didn't list properties $\text{(i)-(iii)}$ because I understood them already.

Proof (of property $\text{(iv)}$). We prove $\text{(7)}$ by induction on $|\alpha|$. Suppose first $|\alpha|=1$. Choose any $\phi \in C_c^\infty (U)$. Then 
  \begin{align}
\int_U \zeta u D^\alpha \phi \, dx &= \int_U u D^\alpha (\zeta \phi) - u(D^\alpha \zeta) \phi \, dx \\
&= - \int_U (\zeta D^\alpha u + u D^\alpha \zeta) \phi \, dx
\end{align}
  Thus $D^\alpha (\zeta u)=\zeta D^\alpha u + u D^\alpha \zeta$, as required.
$\quad$ Next assume $l < k$ and formula $\text{(7)}$ is valid for all $|\alpha| \le l$ and all functions $\zeta$. Choose a multiindex $\alpha$ with $|\alpha| = l+1$. Then $\alpha = \beta + \gamma$ for some $|\beta|=l, |\gamma| = 1$. Then for $\phi$ as above,
\begin{align}
\int_U \zeta u D^\alpha \phi \, dx &= \int_U \zeta u D^\beta (D^\gamma \phi) \, dx \\
&= (-1)^{|\beta|} \int_U \sum_{\sigma \le \beta} {\beta \choose \sigma} D^\sigma \zeta D^{\beta - \sigma} u D^\gamma \phi \, dx  \tag{A} \\
&= (-1)^{|\beta|+|\gamma|} \int_U \sum_{\sigma \le \beta} {\beta \choose \sigma} D^\gamma(D^\sigma \zeta D^{\beta - \sigma} u ) \phi \, dx \tag{B} \\
&= (-1)^{|\alpha|} \int_U \sum_{\sigma\le \beta} {\beta \choose \sigma} [D^\rho \zeta D^{\alpha - \rho} u + D^\sigma \zeta D^{\alpha - \sigma} u] \phi \, dx \tag{C} \\
&= (-1)^{|\alpha|} \int_u \left[\sum_{\sigma \le \alpha} {\alpha \choose \sigma} D^\sigma \zeta D^{\alpha - \sigma} u  \right] \phi \, dx. \tag{D}
\end{align}
  
  
*
  
*$\text{(A)}$: by the induction assumption 
  
*$\text{(B)}$: by the induction assumption again
  
*$\text{(C)}$: where $\rho = \sigma + \gamma$
  
*$\text{(D)}$: since $\displaystyle {\beta \choose \sigma-\gamma} + {\beta \choose \sigma} = { \alpha \choose \sigma}$
  

I especially would like to know how the last step of the work was derived, that is, the line with the tag $\text{(D)}$. I have trouble filling in the details there.
 A: Step (B) isn't really by the induction hypothesis, it's the definition of weak $D^\gamma$ (a.k.a., "formal integration by parts").  
Step (C) is by the induction hypothesis, distributing $D^\gamma$ according to the Leibniz rule.  
To understand (D), split the sum in (C) in two, express the first one in terms of $\rho$, and  then rename the index $\rho$ as $\sigma$:
$$\sum_{\sigma\le \beta} {\beta \choose \sigma} D^\rho \zeta D^{\alpha - \rho} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u \\
= \sum_{\gamma \le \rho\le \alpha} {\beta \choose \rho-\gamma} D^\rho \zeta D^{\alpha - \rho} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u \\
= \sum_{\gamma\le \sigma\le \alpha} {\beta \choose \sigma-\gamma} D^\sigma \zeta D^{\alpha - \sigma} u + \sum_{\sigma\le \beta} {\beta \choose \sigma}D^\sigma \zeta D^{\alpha - \sigma} u $$
It remains to use the aforementioned identity 
$\displaystyle {\beta \choose \sigma-\gamma} + {\beta \choose \sigma} = { \alpha \choose \sigma}$, which can be proved by recalling that multinomial coefficient ${ \alpha \choose \sigma}$ is the coefficient of $x^\sigma$ in $$(1+x)^\alpha = (1+x)^\beta(1+x)^\gamma = (1+x)^\beta + x^\gamma (1+x)^\beta$$
Here I am (ab)using notation, how it's customary with multiindices: e.g., $$(1+x)^\alpha = \prod_i (1+x_i)^{\alpha_i}$$  Since $|\gamma|=1$, the factor $(1+x)^\gamma$ is linear: it's simply $1+x_i$ where $i$ is whatever coordinate has $\gamma_i=1$.
