calculus - Taylor Expansion for a Multivariable Function

\begin{align}
f(x_1,\dots,x_d) &= \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_d = 0}^\infty
\frac{(x_1-a_1)^{n_1}\cdots (x_d-a_d)^{n_d}}{n_1!\cdots n_d!}\,\left(\frac{\partial^{n_1 + \cdots + n_d}f}{\partial x_1^{n_1}\cdots \partial x_d^{n_d}}\right)(a_1,\dots,a_d) \\
&= f(a_1, \dots,a_d) \\
&\quad + \sum_{j=1}^d \frac{\partial f(a_1, \dots,a_d)}{\partial x_j} (x_j - a_j) \\
&\quad + \sum_{j=1}^d \sum_{k=1}^d \frac{1}{2!} \frac{\partial^2 f(a_1, \dots,a_d)}{\partial x_j \partial x_k} (x_j - a_j)(x_k - a_k) \\
&\quad + \sum_{j=1}^d\sum_{k=1}^d\sum_{l=1}^d \frac{1}{3!} \frac{\partial^3 f(a_1, \dots,a_d)}{\partial x_j \partial x_k \partial x_l} (x_j - a_j)(x_k - a_k)(x_l - a_l) \\
&\quad + \dots
\end{align}

I have read through wikipedia, and when I saw the above formula, I didn't know how the second equality is justified. Anybody can help me please? (The first equality is assumed to be true by myself thus doesn't need to be proved)

Answer

The second RHS is an enumeration of the first RHS according to the value of $m=n_1+\cdots+n_d$. For $m=0$, one gets one term, which is $f(a_1, \dots,a_d)$. For $m=1$, one gets $d$ terms, which are the products $\frac{\partial f(a_1, \dots,a_d)}{\partial x_j}\cdot(x_j - a_j)$ for each $1\leqslant j\leqslant d$.
More generally, for each $m\geqslant0$, one gets $d^m$ terms, hence the multiple sums from $1$ to $d$ with $m$ sums.

To "sum" the above, one uses the identity
$$
\sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_d = 0}^\infty A(n_1,\cdots,n_d) =\sum_{m=0}^\infty\sum_{\begin{array}{c}(n_1,\cdots,n_d)\\ n_1+\cdots+n_d=m\end{array}} A(n_1,\cdots,n_d),
$$
with

$$
A(n_1,\cdots,n_d)=\frac{\partial^m f(a_1, \dots,a_d)}{\partial^{n_1} x_1\cdots \partial^{n_d} x_d}\cdot\prod_{j=1}^m (x_j - a_j)^{n_j}.
$$