A proof for the Taylor's Theorem Lagrange with Lagrange remainder in several variables. Theorem (Taylor's Theorem-Lagrange): Let $U \subset \mathbb{R}^m$ a open and $f: U \to \mathbb{R}$ a function of class $C^k$. If the segment $[a,a+v] \subset U$ and $f$ $k+1$ times differentiable in the open $(a,a+v)$, then there exits $\theta \in (0,1)$, such that
$$f(a+v) = f(a) + f'(a)\cdot v + \frac{1}{2!}f''(a) \cdot v^2 + \cdots + \frac{1}{k!}f^{(k)}(a)\cdot v^k + r(v),$$
where $r(v) = \frac{1}{(k+1)!}f^{(k+1)}(a+\theta v) \cdot v^{k+1}$ and $v^k = (v, \ldots, v)$.
I've already seen the proof for the context of real functions, using the Mean Value Theorem, but I'm not able to adapt it to the stated context. I would like to make a proof that does not depend on the case in a variable, which is direct. From what I researched, I would have to define $\psi: [0,1] \to \mathbb{R}$ by $\psi(t) = f(a+tv)$, but I'm not able to make the proper adaptation.
If you can tell me some references about this subject too, let it be done in several variables.
Thank you for your help.
 A: We first make a convention of dot products.
Convention. Without defining $v^n$ or $f^{(n)}(a)$, we define
$$f^{(n)}(a) \cdot v^n := \sum\limits_{i_n=1}^m\sum\limits_{i_{n-1}=1}^m...\sum\limits_{i_1=1}^m \frac{\partial^nf}{\partial x_{i_1}\partial x_{i_2}...\partial x_{i_n}}(a)v_{i_1}v_{i_2}...v_{i_n}.$$
$f$ is a map from $U \subseteq \mathbb R^m$ to $\mathbb R$ hence $\frac{\partial f}{\partial x_{i_1}}$ is, too($\frac{\partial f}{\partial x_{i_1}}$ has $U$ as its domain and the codomain is $\mathbb R$). In a similar manner the whole $\frac{\partial^nf}{\partial x_{i_1}\partial x_{i_2}...\partial x_{i_n}}(a)$ is a scalar in $\mathbb R$, and so is the whole RHS equation.
Noted that $f^{(n)}(a) \cdot v^n = \underbrace{ [(...(\nabla(\nabla f \cdot v) \cdot v)...) \cdot v]}_{n \text{ times of } \nabla}(a)$ as you should verify. For instance
$$ f'(a) \cdot v = [\nabla f \cdot v](a) = \sum\limits_{j=1}^m \frac{\partial f}{\partial x_j}(a)v_j$$
and
$$ f''(a) \cdot v^2 = [\nabla (\nabla f \cdot v) \cdot v](a) = \sum\limits_{k=1}^m\sum\limits_{j=1}^m \frac{\partial^2 f}{\partial x_j \partial x_k}(a)v_jv_k.$$
Play around with some examples would help a lot to understand this.

Recall that:
$$ g(1) = g(0) + g'(0)(1-0) + ... + \frac{1}{(n-1)!}g^{(n-1)}(0)(1-0)^{n-1} + r(v)$$
where $r(v) = \frac{1}{n!}g^{(n)}(v)(1-0)^k$ for some $r \in [0,1]$. We choose $g(x) = f(a(1-t) + bt)$ where $x \in [0,1]$, then by chain rule of multivariable function, $\frac{df}{dt}(f(a(1-t) + bt)) = (\nabla f)^{T}(b-a) = \nabla f \cdot (b-a) $; hence:
$$ f(b) = f(a) + \nabla f(a) \cdot (b-a) + \nabla [\nabla f \cdot (b-a)](a)\cdot (b-a)+...$$
$$ + \frac{1}{(n-1)!}\underbrace{[(...(\nabla(\nabla f \cdot (b-a)) \cdot (b-a))...)(a) \cdot (b-a)]}_{n-1 \text{ times of } \nabla} + R(v) $$
where $R(v) = \underbrace{[(...(\nabla(\nabla f \cdot (v)) \cdot (v))...)(a) \cdot v]}_{n \text{ times of } \nabla}$ for some $v = a(1-t)+bt$, $t \in [0,1]$.
By our convention,
$$ f(b) = f(a) + f'(a) \cdot(b-a) + ... + \frac{1}{(n-1)!}f^{(n-1)}(a) \cdot (b-a)^{n-1} + R(b-a).$$
Or to say
$$ f(a+v) = f(a) + f'(a) \cdot(v) + ... + \frac{1}{(n-1)!}f^{(n-1)}(a) \cdot (v)^{n-1} + R(v).$$

Note: $\nabla[\nabla f \cdot v]$ is the derivative of $\nabla f \cdot v$ with respect to the input of $\nabla f$.
Also, see about tensor & https://en.wikipedia.org/wiki/Dyadics if you shall know what $v^n$ or $f^{n}(a)$ means by itself. But it is the same thing just in a different language.
