Proof of multi-index Taylor formula Assume that $f:\mathbb{R}^N \to \mathbb{R}$ is smooth. 
Where can I find an elementary proof of the multi-index Taylor formula?

$$f(x+h) = \sum_{|\alpha|\le k} \frac{1}{\alpha !} D^\alpha f(x)h^\alpha + O(|h|^{k+1}) \quad \text{ as } h \to 0$$
  for each $k = 1,2,\dots$

 A: Take $g(t) = f(x+th)$. By the Taylor formula for $1$-variable we have:
$$g(t) = \sum_{j=0}^k \frac{1}{j!}g^{(j)}(0) t^j+ \frac{1}{k!}\int_0^1 (t-s)^{k}g^{(k+1)}(s)ds.$$
So
$$f(x+h) = g(1) = \sum_{j=0}^k \frac{1}{j!}g^{(j)}(0) + \frac{1}{k!}\int_0^1 (1-s)^{k}g^{(k+1)}(s)ds.$$
Notice that
$$g'(t) = \sum_{i=1}^n\partial_i f(x+th)h_i = h\nabla f(x+th)$$
where $h\nabla = \sum_{i=1}^{n} h_i\partial_i$ is a differential operator.
The second derivative will be
$$g''(t) = \frac{d}{dt} h\nabla f(x+th)= h\nabla \frac{d}{dt} f(x+th) = h\nabla \lbrace h\nabla f(x+th)\rbrace  = (h\nabla)^2 f(x+th).$$
By induction we have
$$g^{(j)}(t) = (h\nabla)^j f(x+th).$$
By the binomial formula we have:
$$(h\nabla)^j = \left(\sum_{i=1}^{n} h_i\partial_i\right)^j = \sum_{|\alpha| = j} \frac{j!}{\alpha!} h^\alpha \partial^\alpha.$$
Therefore $$g^{(j)}(t) =(h\nabla)^j f(x+th) =\sum_{|\alpha| = j} \frac{j!}{\alpha!} h^\alpha \partial^\alpha f(x+th).$$
Now just plug the formula above on the one dimensional Taylor and you obtain:
$$f(x+h) = \sum_{j=0}^k \frac{1}{j!}\sum_{|\alpha| = j} \frac{j!}{\alpha!} h^\alpha \partial^\alpha f(x)+$$  $$+\frac{1}{k!}\int_0^1 (1-s)^{k}\sum_{|\alpha| = k+1} \frac{(k+1)!}{\alpha!} h^\alpha \partial^\alpha f(x+th).$$
Simplifying the above expression you obtain
$$f(x+h) = \sum_{j=0}^k \sum_{|\alpha| = j}\frac{1}{\alpha!} h^\alpha \partial^\alpha f(x)+\sum_{|\alpha| = k+1}\frac{k+1}{\alpha!}\int_0^1 (1-s)^{k} h^\alpha \partial^\alpha f(x+sh)ds.$$
Now just observe that $\sum_{j=0}^k \sum_{|\alpha| = j} = \sum_{|\alpha|\leq k}$ and we obtain:
$$f(x+h) = \sum_{|\alpha| \leq k}\frac{1}{\alpha!} h^\alpha \partial^\alpha f(x)+\sum_{|\alpha| = k+1}\frac{k+1}{\alpha!}\int_0^1 (1-s)^{k} h^\alpha \partial^\alpha f(x+sh)ds.$$
That is essentially what you want. It is easy to see that the remainder is $O(|h|^{k+1})$ because $\partial^\alpha f(x+th)$ is continuous for all $|\alpha| = k+1$.
