Prove: ∫0∞x2ne−x2dx=(2n)!22nn!π2

Alternatively, set
${\displaystyle I\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{e}^{-a{x}^2}dx}$
differentiate n times with respect to ${\displaystyle \alpha }$ and evaluate at ${\displaystyle \alpha =1}$
To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that ${\displaystyle I\left(\alpha \right)=\frac{1}{2}\sqrt{\frac{\pi }{\alpha }}}$ and differentiating to obtain
${\displaystyle \frac{d^n}{d{a}^n}I\left(a\right)={(-1)}^n{\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-a{x}^2}dx}$
some algebraic manipulation and evaluating at ${\displaystyle a=1}$ will yield the wanted identity.

Alternatively, integration by parts works immediately.
${\displaystyle a_n={\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-x^2}}$
Consider ${\displaystyle U=x^{2n-1}}$ so that ${\displaystyle du=(2n-1)x^{2n-2}}$, and ${\displaystyle dv=x{e}^{-x^2}}$ so that ${\displaystyle V=-\frac{1}{2}{e}^{-x^2}}$.
Then
${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-x^2}dx=\frac{1}{2}{e}^{-x^2}{x}^{2n-1}{\mid }_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}(2n-1)x^{2n-2}\frac{-1}{2}{e}^{-x^2}dx}$
${\displaystyle =\frac{2n-1}{2}{\int }_0^{\mathrm{\infty }}{x}^{2n-2}{e}^{-x^2}dx=\frac{(2n-1)2n}{2^{2}n}{\int }_0^{\mathrm{\infty }}{x}^{2n-2}{e}^{-x^2}dx}$
Hence
${\displaystyle a_n=\frac{\left(2n\right)(2n-1)}{2^{2}n}{a}_{n-1}}$
and since ${\displaystyle a_0=\frac{\sqrt{\pi }}{2}}$ we conclude
${\displaystyle a_n={\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-x^2}dx=\frac{\left(2n\right)!}{2^{2n}n!}\frac{\sqrt{\pi }}{2}}$

Let's suppose that, one way or another, you know that ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$. Then
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2tx-x^2}dx=e^{t^2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(t-x{)}^2}dx=e^{t^2}\sqrt{\pi }$
On the other hand,
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2tx-x^2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(\sum _{n\ge 0}\frac{2^n{t}^n{x}^n}{n!})e^{-x^2}dx=\sum _{n\ge 0}\frac{2^n{t}^n}{n!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^n{e}^{-x^2}dx$
Finally, note that by evenness,
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2n}{e}^{-x^2}dx=2{\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-x^2}dx$