Let $\alpha=(\alpha_1,\dots \alpha_n)$ be a multi-index of order $\lvert \alpha \rvert=\alpha_1+\cdots+\alpha_n $ and let $u\colon\mathbb{R}^n\to \mathbb{R}$ be a sufficiently regular function. I must prove that
$$\large\frac{1}{|\alpha|!}d^{|\alpha|}u(x)=\sum_{|\alpha|=k}\frac{1}{\alpha !}D^\alpha u(x)(dx)^\alpha.$$ where $d^{|\alpha|}u$ denotes the differential of order $|\alpha|$ of $u$. My idea is to use the multinomial theorem, but I don't know if what I write is formally correct:
$$\begin{align} \frac{1}{|\alpha|!}d^{|\alpha|} u(x)\color{BLUE}{=}\frac{1}{|\alpha|!}&\left(\frac{\partial}{\partial x_1}dx_1+\cdots+\frac{\partial}{\partial x_n}dx_n\right)^{|\alpha|}u\\ \color{RED}{=}& \frac{1}{|\alpha|!}\left(\sum_{|\alpha|=k}\frac{|\alpha|!}{\alpha!}\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}(d{x_1})^{\alpha_1}\cdots\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}(d{x_n})^{\alpha_n}\right)u\\ =& \sum_{|\alpha|=k}\frac{1}{\alpha !}\frac{\partial^{|\alpha|}u}{\partial x_1^{\alpha_1}\cdots \partial x_n^{\alpha_n}}(d{x_1})^{\alpha_1}\cdots (dx_n)^{\alpha_n} \\ =& \sum_{|\alpha|=k} \frac{1}{\alpha !} D^\alpha u(x)(dx)^\alpha \end{align} $$
In the equality in blue I used the definition of differential and in the equality in red I used the multinomial theorem.
I seem to write the following equality:
$$\left(\frac{\partial}{\partial x_i}\right)^{\alpha_i}=\frac{\partial^{\alpha_i}}{\partial x_i^{\alpha_i}}$$
Question What's right and what's wrong? Is my proof formally correct?