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!
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 $\int_{-\infty}^\infty e^{\pi t^2}\,dt=1$