Fourier transform 4 times = original function (from Bracewell book)
I'll use the non-unitary Fourier transform (but this is not important, it's just a preference):
$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$
$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega\tag{2}$$
where (1) is the Fourier transform, and (2) is the inverse Fourier transform.
Now if you formally take the Fourier transform of $X(\omega)$ you get
$$\mathcal{F}\{X(\omega)\}=\mathcal{F}^2\{x(t)\}=\int_{-\infty}^{\infty}X(\omega)e^{-i\omega t}d\omega\tag{3}$$
Comparing (3) with (2) we have
$$\mathcal{F}^2\{x(t)\}=2\pi x(-t)\tag{4}$$
So the Fourier transform equals an inverse Fourier transform with a sign change of the independent variable (apart from a scale factor due to the use of the non-unitary Fourier transform).
Since the Fourier transform of $x(-t)$ equals $X(-\omega)$, the Fourier transform of (4) is
$$\mathcal{F}^3\{x(t)\}=2\pi X(-\omega)\tag{5}$$
And, by an argument similar to the one used in (3) and (4), the Fourier transform of $X(-\omega)$ equals $2\pi x(t)$. So we obtain for the Fourier transform of (5)
$$\mathcal{F}^4\{x(t)\}=2\pi\mathcal{F}\{X(-\omega)\}=(2\pi)^2x(t)\tag{6}$$
which is the desired result. Note that the factor $(2\pi)^2$ in (6) is a consequence of using the non-unitary Fourier transform. If you use the unitary Fourier transform (where both the transform and its inverse get a factor $1/\sqrt{2\pi}$) this factor would disappear.
In sum, apart from irrelevant constant factors, you get
$$\bbox[#f8f1ea, 0.6em, border: 0.15em solid #fd8105]{x(t)\overset{\mathcal{F}}{\Longrightarrow} X(\omega)\overset{\mathcal{F}}{\Longrightarrow} x(-t)\overset{\mathcal{F}}{\Longrightarrow} X(-\omega)\overset{\mathcal{F}}{\Longrightarrow} x(t)}$$