If $f(x)= e^{x^2}$ show that $f^{(2n)}(0)= (2n)!/ n!$.
The power series for $e^{x^2}$ is $\sum_{k=0}^\infty\frac{x^{2k}}{k!}$ (just substitute $x^2$ for x in the usual power series for $e^x$). Compare that with the Taylor series formula $f(x) = \sum_{r=0}^\infty\frac{f^{(r)}(0)}{r!}x^r$, and you see that, for the function $f(x) = e^{x^2}$, $f^{(r)}(0) = 0$ if r is odd. If $r=2k$ is even, then compare the coefficients of $x^{2k}$ in the two series to see that $\frac1{k!} = \frac{f^{(2k)}(0)}{(2k)!}$.