Theorem If $i,k$ are multi-indices in $\sN^n$ , and $x=(x_1,\ldots, x_n)$ , then \begin{eqnarray*} \partial^i x^k = \left\{ \begin {array}{ll} \frac{k!}{(k-i)!} x^{k-i} & \mbox{if}\,\, i\le k, \\ 0 & \mbox{otherwise}. \end{array}.

Proof. The proof follows from the corresponding rule for the ordinary derivative; if $i,k$ are in $0,1,2,\ldots$ , then \begin{eqnarray} \label{usual} \frac{d^i}{dx^i} x^k = \left\{ \begin {array}{ll} \frac{k!}{(k-i)!} x^{k-i} & \mbox{if}\,\, i\le k, \\ 0 & \mbox{otherwise.} \end{array}. Suppose $i=(i_1,\ldots, i_n)$ , $k=(k_1,\ldots, k_n)$ , and $x=(x_1,\ldots, x_n)$ . Then we have that \begin{eqnarray*} \partial^i x^k &=& \frac{\partial^{|i|}}{\partial x_1^{i_1} \cdots \partial x_n^{i_n}} x_1^{k_1} \cdots x_n^{k_n} \\ &=& \frac{\partial^{i_1}}{\partial x_1^{i_1}} x_1^{k_1} \cdot \cdots \cdot \frac{\partial^{i_n}}{\partial x_n^{i_n}} x_n^{k_n}. \end{eqnarray*}For each $r=1,\ldots, n$ , the function $x_r^{k_r}$ only depends on $x_r$ . In the above, each partial differentiation $\partial/\partial x_r$ therefore reduces to the corresponding ordinary differentiation $d/dx_r$ . Hence, from equation , it follows that $\partial^i x^k$ vanishes if $i_r > k_r$ for any $r=1,\ldots, n$ . If this is not the case, i.e., if $i\le k$ as multi-indices, then for each $r$ , $$\frac{d^{i_r}}{dx_r^{i_r}} x_r^{k_r} = \frac{k_r!}{(k_r-i_r)!} x_r^{k_r-i_r},$$ and the theorem follows.