Editing Multi-index notation (section)

===Proof===
The proof follows from the [[power rule]] for the [[differential calculus|ordinary derivative]]; if ''α'' and ''&beta;'' are in <math display="inline">\{0, 1, 2,\ldots\}</math>, then
{{NumBlk||<math display="block"> \frac{d^\alpha}{dx^\alpha} x^\beta = \begin{cases}
\frac{\beta!}{(\beta-\alpha)!} x^{\beta-\alpha} & \hbox{if}\,\, \alpha\le\beta, \\
0 & \hbox{otherwise.}
\end{cases}</math>|{{EquationRef|1}}}}

Suppose <math>\alpha=(\alpha_1,\ldots, \alpha_n)</math>, <math>\beta=(\beta_1,\ldots, \beta_n)</math>, and <math>x=(x_1,\ldots, x_n)</math>. Then we have that
<math display="block">\begin{align}\partial^\alpha x^\beta&= \frac{\partial^{\vert\alpha\vert}}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n}} x_1^{\beta_1} \cdots x_n^{\beta_n}\\
&= \frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}} x_1^{\beta_1} \cdots
\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}} x_n^{\beta_n}.\end{align}</math>

For each <math display="inline">i</math> in <math display="inline">\{ 1, \ldots , n\}</math>, the function <math>x_i^{\beta_i}</math> only depends on <math>x_i</math>. In the above, each partial differentiation <math>\partial/\partial x_i</math> therefore reduces to the corresponding ordinary differentiation <math>d/dx_i</math>. Hence, from equation ({{EquationNote|1}}), it follows that <math>\partial^\alpha x^\beta</math> vanishes if <math display="inline">\alpha_i > \beta_i</math> for at least one <math display="inline">i</math> in <math display="inline">\{ 1, \ldots , n\}</math>. If this is not the case, i.e., if <math display="inline">\alpha \leq \beta</math> as multi-indices, then
<math display="block"> \frac{d^{\alpha_i}}{dx_i^{\alpha_i}} x_i^{\beta_i} = \frac{\beta_i!}{(\beta_i-\alpha_i)!} x_i^{\beta_i-\alpha_i}</math>
for each <math>i</math> and the theorem follows. [[Q.E.D.]]