$$ \int_{-\infty}^{+\infty}{\rm d}x~e^{x^2}\left(\frac{{\rm d}^n e^{-x^2}}{{\rm d}x^n}\right)^2 = 2^n n!\sqrt{\pi} $$
For the last two weeks, I have been unable to prove this integral which I came across. I tried converting it to a double integral and doing a change of variables, I tried turning it into a series and I attempted integration by parts but I somehow could not prove the integral.
How does one prove the integral shown in the picture above?
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} \int_{-\infty}^{\infty}\expo{x^{2}}\pars{\totald[n]{\expo{-x^{2}}}{x}}^{2} \,\dd x & = \left.\partiald[n]{}{a}\partiald[n]{}{b}\int_{-\infty}^{\infty}\expo{x^{2}} \expo{-\pars{x - a}^{2}}\expo{-\pars{x - b}^{2}} \,\dd x\,\right\vert_{\ a\ =\ b\ =\ 0}\label{1}\tag{1} \\[5mm] & = \left.\partiald[n]{}{a}\partiald[n]{\pars{\root{\pi}\expo{2ab}}}{b} \right\vert_{\ a\ =\ b\ =\ 0} \\[5mm] & = \left.\root{\pi}\,\partiald[n]{\bracks{\pars{2a}^{n}\expo{2ab}}}{a} \right\vert_{\ a\ =\ b\ =\ 0} \\[5mm] & = \left.2^{n}\root{\pi}\braces{n!\bracks{a^{n}}\pars{a^{n}\expo{2ab}}} \right\vert_{\ a\ =\ b\ =\ 0} \\[5mm] & = \left.2^{n}\,n!\root{\pi}\bracks{a^{0}}\expo{2ab}\right\vert_{\ a\ =\ b\ =\ 0} = \bbx{2^{n}\,n!\root{\pi}} \end{align}