Fourier transform of the Fourier transform Is the Fourier transform of the Fourier transform of $f(t)$: $$\hat{\hat{f(t)}} = f(-t)$$ or  $$\hat{\hat{f(t)}} = 2\pi f(-t)$$
?
I have read the two versions here and here (respectively) for example.
And also, can I show this from the definition ? I tried this :
$$ \hat{\hat{f(t)}} = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\right)e^{-i\omega t}dt $$ 
$$= \hat{f(\omega)}\int_{-\infty}^{\infty}e^{-i\omega t}dt$$
but I don't really know where I'm going with this...
 A: UPDATE:
The proof below might be wrong, due to a faulty substituion, as pointed out by Dan in the comments. 
I marked the equation in question with a question mark.
Original Proof Attempt:
You reuse the $t$ and $\omega$ variables, also you factor out $\hat{f}(\omega)$ while there is an integration running over it and it is not constant. So your derivation might better start like this:
\begin{align}
\hat{\hat{f}}(\omega)
&= 
\int\limits_{-\infty}^{+\infty}\hat{f}(t)e^{-i\omega t}dt \\
&= 
\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
f(\tau)e^{-it \tau}d\tau \, e^{-i\omega t}dt \\
&= 
\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
f(\tau)e^{-it (\tau+\omega)}d\tau \, dt \\
\end{align}
Switching integration order gives
$$
\begin{align}
\hat{\hat{f}}(\omega)
&=
\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
f(\tau)e^{-it (\tau+\omega)}dt \, d\tau \\
&= 
\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
e^{-it (\tau+\omega)}dt \, f(\tau) \, d\tau \\
&{? \atop =} 
\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
e^{-it u}du \, f(\tau) \, d\tau \\
&= \int\limits_{-\infty}^{+\infty} \!\!
2\pi\delta(u) \, f(\tau) d\tau \\
&= 2\pi \int\limits_{-\infty}^{+\infty} \!\!
\delta(\tau + \omega) \, f(\tau) d\tau \\
&= 2\pi \int\limits_{-\infty}^{+\infty} \!\!
f(\tau) \, \delta(\tau-(-\omega)) \, d\tau \\
&= 
2\pi \, f(-\omega)
\end{align}
$$
where in between a substitution $u = t + \omega$ with $du = dt$ was used.
