Answer
${f^{2n}(0)}=\frac{(2n)!}{n!}$
Work Step by Step
${e^{x}}=\Sigma_{n=0}^\infty\frac{x^{n}}{n!}$
$f(x)={e^{x^{2}}}=\Sigma_{n=0}^\infty\frac{(x^{2})^{n}}{n!}=\Sigma_{n=0}^\infty\frac{x^{2n}}{n!}$
We can see that the coefficient of $x^{2n}$ in the power series of $f(x)$ is $\frac{1}{n!}$.
Thus,
$\frac{f^{2n}(0)}{2n!}=\frac{1}{n!}$
Hence, ${f^{2n}(0)}=\frac{(2n)!}{n!}$
