Computing the Fourier transform of $H_k(x)e^{-x^2/2}$, where $H_k$ is the Hermite polynomial.

Yes, once you have the equation $$ \sum_{n}\mathscr{F}(e^{-x^2/2}H_n(x))\frac{t^n}{n!} = \sum_{n}\sqrt{2\pi}(-i)^n H_n(\xi)e^{-\xi^2/2}\frac{t^n}{n!}, $$ then you can fix $\xi$, and view this as a power series equation that holds for all $t$. Therefore, for this fixed $\xi$, the power series coefficients must be identical, which leads to $$ \mathscr{F}(e^{-x^2}H_n(x))=\sqrt{2\pi}(-i)^n H_n(\xi) $$ The left side implicitly depends on $\xi$, because it is the transform variable. The transform is continuous in $\xi$ because of the exponentially decaying nature of the function being transformed. So $$ \mathscr{F}(e^{-x^2}H_n(x))(\xi)=\sqrt{2\pi}(-i)^n H_n(\xi),\;\;\xi\in\mathbb{R}. $$ Omitting the arguments of the functions, these functions are equal: $$ \mathscr{F}(e^{-x^2}H_n(x)) = \sqrt{2\pi}(-i)^n H_n $$