Showing that the Fourier transform of $f_n(t)=e^{\pi t^2}\frac{d^n}{dt^n}e^{-2\pi t^2}$ is $\hat{f_n} = (-i)^nf_n$ Let $n \in \mathbb{N}_0$ and define $f_n(t)=e^{\pi t^2}\frac{d^n}{dt^n}e^{-2\pi t^2}$. I am tasked to show that $\hat{f_n} = (-i)^nf_n$ for $\hat{f_n}$ the Fourier transform of $f_n$. My problem is that I am either not seeing the integral trick I am supposed to use or I fail to see the pattern in the derivatives of $e^{-2\pi t^2}$ as I don't know how to proceed beyond
$$\hat{f_n}(y) = \int_{\mathbb{R}}\left(e^{\pi t^2}\frac{d^n}{dt^n}e^{-2\pi t^2}\right)e^{-2\pi i t y}dt$$
I am not looking for a complete solution per se, but rather a better intuition for this problem. Although if you happen to know some elegant and general method for dealing with this problem, feel free to share it! Thanks!
 A: Note that we have
$$\begin{align}
\int_{-\infty}^\infty e^{\pi t^2-i2\pi yt}\frac{d^n}{dt^n}e^{-2\pi t^2}\,dt&=e^{\pi y^2}\int_{-\infty}^\infty e^{\pi(t-iy)^2}\frac{d^n}{dt^n}e^{-2\pi t^2}\,dt\tag1\\\\
&=e^{\pi y^2}\int_{-\infty-iy}^{\infty-iy} e^{\pi t^2}\frac{d^n}{dt^n}e^{-2\pi (t+iy)^2}\,dt\tag2\\\\
&=e^{\pi y^2}\int_{-\infty-iy}^{\infty-iy} e^{\pi t^2}(-i)^n\frac{d^n}{dy^n}e^{-2\pi (t+iy)^2}\,dt\tag3\\\\
&=(-i)^ne^{\pi y^2}\frac{d^n}{dy^n}\int_{-\infty-iy}^{\infty-iy} e^{\pi t^2}e^{-2\pi (t+iy)^2}\,dt\tag4\\\\
&=(-i)^ne^{\pi y^2}\frac{d^n}{dy^n}\left(e^{2\pi y^2}\int_{-\infty-iy}^{\infty-iy}e^{-\pi t^2-i4\pi yt} \,dt\right)\tag5\\\\
&=(-i)^ne^{\pi y^2}\frac{d^n}{dy^n}\left(e^{-2\pi y^2}\underbrace{\int_{-\infty-iy}^{\infty-iy}e^{-\pi (t-i2 y)^2} \,dt}_{=1}\right)\tag 6\\\\
&=(-i)^ne^{\pi y^2}\frac{d^n}{dy^n}e^{-2\pi y^2}
\end{align}$$
as was to be shown!

NOTES:
$(1)$ Completed the square in the exponent of $e^{\pi t^2-i2\pi yt}$ and factored out the term $e^{\pi y^2}$
$(2)$ Enforced the substitution $t\mapsto t+iy$
$(3)$ Exploited the relationship $\frac{de^{-2\pi (t+iy)}}{dt}=(-i)\frac{de^{-2\pi (t+iy)}}{dy}$, $n$ times
$(4)$ Interchanged the integral with the differential operator on $y$
$(5)$ Expanded the square of the exponential $e^{-2\pi (t+iy)^2}$ and factored out the term $e^{2\pi y^2}$
$(6)$  Completed the square, factored out the exponential $e^{-4\pi y^2}$ and combined with the term $e^{2\pi y^2}$, exploited Cauchy's Integral Theorem to deform the contour back to the real axis, and made use of the identity $\int_{-\infty}^\infty e^{\pi t^2}\,dt=1$
