Fractional derivative according to Fourier

Question

I am trying to show that $\frac{d^\alpha}{dx^\alpha}e^{ikx}=(ik)^\alpha e^{ikx}$ when using the following equations for fourier transforms: \begin{align} g(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{ikx}dx \\ \frac{d^\alpha}{dx^\alpha}f(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)(-ik)^\alpha e^{-ikx}dk \end{align} Since in this case $f(x)=e^{ikx}$, $g(k)$ would be: \begin{equation} g(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ikx}e^{ikx}dx \end{equation} I can't calculate $g(k)$, does anyone have any ideia on how i can do it?