Answer
The detailed proof is below.
Work Step by Step
$$\frac{d}{dx}(x^{-m})=\frac{d}{dx}\Big(\frac{1}{x^m}\Big)$$
Apply the Derivative Quotient Rule here:
$$\frac{d}{dx}(x^{-m})=\frac{(1)'x^m-1(x^m)'}{(x^m)^2}$$
$$\frac{d}{dx}(x^{-m})=\frac{0\times x^m-mx^{m-1}}{x^{2m}}$$
$$\frac{d}{dx}(x^{-m})=\frac{-mx^{m-1}}{x^{2m}}$$
$$\frac{d}{dx}(x^{-m})=-mx^{m-1-2m}=-mx^{-m-1}$$
The Power Rule for negative integers has been proved.