Multiplicities of Polynomials I'm trying to understand the following explanation. This comes up as a first step into defining intersection multiplicities for Bezout's theorem.

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 $Q$.

Specifically, I'm trying to comprehend the last line. If I recall correctly, a taylor expansion of a multivariate polynomial at point $a$ is given $$ P(x) = \sum_\alpha \frac{\partial^\alpha P(a)}{\alpha !} (x-a)^\alpha.$$
According to the above explanation, $$\frac{\partial^\alpha P(a)}{\alpha !} (x-a)^\alpha \frac{\partial^\beta Q(a)}{\beta !} (x-a)^\beta = \frac{\partial^{\alpha+\beta} PQ(a)}{(\alpha+\beta) !} (x-a)^{\alpha+\beta} $$
but I don't get why this is. We have, simply using definitions, that $ (x-a)^{\alpha} (x-a)^{\beta} = (x-a)^{\alpha+\beta}$. But I still don't get why the above expression holds.
Edit: We do NOT have that $\alpha ! \beta != (\alpha+\beta)!$, upon closer reflection.
 A: You have to combine like terms. So when you multiply $P(x)$ and $Q(x)$ you get
$$P(x)Q(x) = \sum_{\gamma} \left[ \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!} \right] (x - a)^\gamma. \tag{1} $$
Now comparing coefficients with the Taylor expansion of $PQ$:
$$ (PQ)(x) = \sum_\gamma \frac{\partial^\gamma (PQ)(a)}{\gamma!} (x - a)^\gamma \tag{2} $$
we get this identity:
$$ \frac{\partial^\gamma (PQ)(a)}{\gamma!} = \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!}. \tag{3} $$

But I like to think about this a different way. The set $\{ (x - a)^\alpha : \alpha \in \mathbb{N}^{n+1}\}$ is a basis for $\mathbb{C}[x_0, \dots, x_n]$. That means there is a unique way of writing any polynomial $P(x)$ as $P(x) = \sum c_\alpha (x - a)^\alpha$. Now, of course we can take derivatives to get the identity
$$ c_\alpha = \frac{\partial^\alpha P(a)}{\alpha!}. \tag{4} $$
And just so everyone's on the same page, this is short for
$$\frac{\partial_{x_0}^{\alpha_0} \cdots \partial_{x_n}^{\alpha_n} P(a_0,\dots,a_n)}{\alpha_0! \cdots \alpha_n!} (x_0 - a_0)^{\alpha_0} \cdots (x_n - a_n)^{\alpha_n}.$$
But then it necessarily follows, that however we write $(PQ)(x)$ in this basis, those must be the coefficients. And remember: these are not power series, so there are no questions of convergence that need to be addressed.
So because $(1)$ is true by multiplication and $(2)$ is true by fact $(4)$, it must follow that $(3)$ is true since we are working with a basis. Notice that this is something you can't do (like this anyways) with power series since $\{ (x - a)^\alpha : \alpha \in \mathbb{N}^{n+1}\}$ is not a basis for the set of power series.
