Given the function $f(x) = e^{-x^2}$, how can I find the $n$-th derivative of $f$ evaluated at $x = 0$, i.e., $f^{(n)}(0)$?

Given the function $f(x) = e^{-x^2}$, how can I find $f^{(n)}(0)$?

The Taylor series for $e^{-x^2}$ can be written by expanding $e^u$ where $u = -x^2$:

$$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$$

This simplifies to:

$$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$$

First of all, why can I simply plug in that $-x^2$and can I just diffrentiate it? but in order to diffrentiate it term by term I need the sum to converge uniformly to my function. Is it tho? I am not really sure what should I do.Any help will be appreciated!