Butterworth High Pass Filter Amplitude Response
If $H_{LP}(j\omega)$ is the frequency response of a Butterworth low-pass filter, then its inverse would go to infinity for $\omega\rightarrow\infty$ (because $H_{LP}(j\omega)$ goes to zero for high frequencies). So that's not what you want.

The correct high-pass frequency response is obtained by replacing $\omega/\omega_c$ by $-\omega_c/\omega$, resulting in

$$|H_{HP}(j\omega)|=\frac{\left|\frac{\omega}{\omega_c}\right|^n}{\sqrt{1+\left(\frac{\omega}{\omega_c}\right)^{2n}}}$$

which approaches unity with increasing frequency, as should be the case for a high-pass filter.