Multiplicities of Polynomials Multiplied This is a follow-up to one of my past questions, Multiplicities of Polynomials. I thought it was more appropriate to ask in a new question. This question is self-contained, I have written the relevant details here.
I'm trying to comprehend the explanation:

Let $P$ be a non-zero polynomial in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right]$ and let $a=\left(a_{0}, \ldots, a_{n}\right)$ be a point in $\mathbb{C}^{n+1}$. The multiplicity $\operatorname{mult}_{a} P$ of $P$ at $a$ is the smallest non-negative integer $m$ such that there exists $\alpha \in \mathbb{N}^{n+1}$ with $|\alpha|=m$ and $\partial^{\alpha} P(a) \neq 0$.
Such a non-negative integer $m$ exists, because, up to some non-zero factors, the partial derivatives $\partial^{\alpha} P(a)$ are the coefficients in the Taylor expansion of $P$ around the point $a .$ Thus if $P$ is non-zero, then at least one of these partial derivatives is different from zero. This argument also shows that mult $_{a} P \leq \operatorname{deg}(P),$ and that $\operatorname{mult}_{a}(P Q)=\operatorname{mult}_{a} P+\operatorname{mult}_{a} Q$ for every non-zero polynomial $Q$ in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right],$ since the Taylor expansion of $P Q$ around $a$ is the product of the Taylor expansions of $P$ and $Q$.

We have the identity $$ \frac{\partial^\gamma (PQ)(a)}{\gamma!} = \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!}. \tag{1} $$
But why does this imply that $\operatorname{mult}_{a}(P Q)=\operatorname{mult}_{a} P+\operatorname{mult}_{a} Q$? The RHS of the equation $(1)$ is a multiple sum, therefore non-zero summands may cancel out to give $0$. In other words, suppose that $\operatorname{mult}_a PQ = r$, $\operatorname{mult}_a P = s$, $\operatorname{mult}_a Q = t$. I get that $r \ge st$, but am not able to conclude that $r = st$.
 A: Thinking in terms of derivatives is making this harder than it needs to be; we are just talking about multiplying polynomials here. Translate the polynomial so that $a = 0$ WLOG. The multiplicity $\text{mult}_0(P)$ is the smallest $m$ for which $P$ has a nonzero term of degree $m$ (meaning multidegree $\alpha$ such that $|\alpha| = m$). Among all the nonzero terms of degree $m$ there is a unique one which is largest in lex order, and the product of this term in $P$ with the corresponding term in $Q$ is the unique nonzero term in $PQ$ of minimal degree which is largest in lex order; an explicit example to make the reasoning clear here is
$$P(x, y) = x^2 + y^2 + \text{higher terms}$$
$$Q(x, y) = x^2 y - y^3 + \text{higher terms}$$
$$P(x, y) Q(x, y) = x^4 y - y^5 + \text{higher terms}$$
where, if we pick the lex order in which $x \ge y$, the lex biggest term in $P$ of minimal degree is $x^2$, the lex biggest term in $Q$ of minimal degree is $x^2 y$, and the lex biggest term in $PQ$ of minimal degree is their product $x^4 y$.
You are correct that there can be cancellation for some $\gamma$, which happens in the example above: $x^2 y (y^2)$ cancels with $- x^2 y^3$. But cancellation doesn't occur for the biggest (or smallest) term in any monomial order because the RHS of your product formula has a single term in this case.
