Evans's proof of the Leibniz's formula for the weak derivatives in Sobolev spaces

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 $\gamma_i=1$.