Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ The problem is stated as:

Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ and $n \in \mathbb{N}$

Attempt:
We know that $e^{x} = \sum_{k=0}^{n} \frac{1}{k!}x^k +\frac{x^{n+1}e^{\theta x}}{(n+1)!}$ for some $\theta = \theta (x) \in (0,1)$
Hence, $e^{x^2}=\sum_{k=0}^{n} \frac{1}{k!}x^{2k} +\frac{x^{2n+1}e^{\theta x^2}}{(n+1)!}$
Comparing terms with the general Maclaurin polynomial, we get that:
$\frac{1}{k!}x^{2k} = \frac{f^{(n)}(0)}{n!}x^{n}$
First, we have to equate the exponent of the $x$ - terms, so $2k = n \iff k = n/2$
Therefore, we have that:
$\Rightarrow f^{n}(0) = \frac{n!}{(n/2)!}$
However, since we had natural numbers in our expression, I wonder what would happen if $n$ was odd in this case. How could I generalize this even further?
Thanks!

You can make the problem more general and compute the $n^{th}$ derivative at any value of $n$.
They all write
$$f^{(n)}(x)=e^{x^2} P_n(x)$$ Computing the first ones (using the chain rule makes it easy)
$$f^{(n+1)}(x)=e^{x^2}\left(2x P_n(x)+P'_n(x) \right)$$
So, you have
$$\left(
\begin{array}{cc}
n & P_n(x) \\
 1 & 2 x \\
 2 & 4 x^2+2 \\
 3 & 8 x^3+12 x \\
 4 & 16 x^4+48 x^2+12 \\
 5 & 32 x^5+160 x^3+120 x \\
 6 & 64 x^6+480 x^4+720 x^2+120 \\
 7 & 128 x^7+1344 x^5+3360 x^3+1680 x \\
 8 & 256 x^8+3584 x^6+13440 x^4+13440 x^2+1680 \\
 9 & 512 x^9+9216 x^7+48384 x^5+80640 x^3+30240 x \\
 10 & 1024 x^{10}+23040 x^8+161280 x^6+403200 x^4+302400 x^2+30240
\end{array}
\right)$$ where you can notice interesting patterns.
In fact
$$\color{red}{P_n=i^{-n} H_n(i x)\implies f^{(n)}(x)=i^{-n} H_n(i x)e^{x^2}}$$
Now, if you make $x=0$, as expected since $e^{x^2}$ is a even function, $f^{(2n+1)}(0)=0$ and for $f^{(2n)}(0)$ generates the  sequence
$$\{2,12,120,1680,30240,665280,\cdots\}$$ which is your result.
