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I have learned the Butterworth filter, normally it is used for low pass design. And the amplitude response is:
$$|H(jw)|=\frac{1}{\sqrt{1+(\frac{w}{w_{c}})^{2n}}}$$
mentioned in this Butterworth Introduction. But as I know, the Butterworth can be designed as a high pass as well. Then what is the amplitude response for high pass? Is it simply like this?
$$|H_{hp}| = \frac{1}{|H(jw)|}$$
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.
