100th derivative of $e^{-x^2}$ at point $0$ Problem: 
Find
$\frac{\mathrm d^{100}}{\mathrm dx^{100}}e^{-x^2}$ at point $0$.
My attempt:
$y'=-2xe^{-x^2}$
I tried to use General Leibniz rule and I didn't get much better information.
 
Without: Taylor series
 A: Since
$$
\begin{align}
\lim_{n\to\infty}\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\frac xn\right)^n
&=\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1}\\
&=e^x
\end{align}
$$
and
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\lim_{n\to\infty}\left(1+\frac xn\right)^n
&=\frac{\mathrm{d}}{\mathrm{d}x}e^x\\
&=e^x
\end{align}
$$
we can interchange derivatives with limits of $\left(1+\frac xn\right)^n$.
Using the Binomial Theorem, we get
$$
\begin{align}
\left.\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{100}\left(1-\frac{x^2}{n}\right)^n\right|_{x=0}
&=\left.\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{100}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{x^{2k}}{n^k}\right|_{x=0}\\
&=\binom{n}{50}\frac{100!}{n^{50}}
\end{align}
$$
Taking the limit, we get
$$
\lim_{n\to\infty}\binom{n}{50}\frac{100!}{n^{50}}=\frac{100!}{50!}
$$
A: Since the most obvious approach is forbidden I propose the following (large) detour:
It is well known that the Fourier transform of a Gaussian is again a Gaussian. E.g., one has
$$\int_{-\infty}^\infty e^{-t^2/2}\cos(\omega t)\>dt=\sqrt{2\pi}e^{-\omega^2/2}\qquad(\omega\in{\mathbb R})\ .$$
Putting $\omega:=x\sqrt{2}$ and substituting $t:=\tau/\sqrt{2}$ leads to
$$f(x):=e^{-x^2}={1\over 2\sqrt{\pi}}\int_{-\infty}^\infty e^{-\tau^2/4}\cos(x\tau)\>d\tau=
{1\over \sqrt{\pi}}\int_0^\infty e^{-\tau^2/4}\cos(x\tau)\>d\tau\ .$$
Since we are working here in the Schwartz space ${\cal S}$ we may differentiate a hundred times under the integral sign and obtain
$$f^{(100)}(x)={1\over \sqrt{\pi}}\int_0^\infty \tau^{100}e^{-\tau^2/4}\cos(x\tau)\>d\tau\ .$$
Putting $x:=0$ here and substituting $\tau:=2\sqrt{u}$ leads to
$$f^{(100)}(0)={2^{100}\over \sqrt{\pi}}\int_0^\infty e^{-u} u^{99/2}\>du={2^{100}\Gamma(101/2)\over\sqrt{\pi}}={100!\over50!}\ .$$
A: You started well..
So you basically got $$y'=-2xy$$
Now we have:
$$y'+2xy=0$$
Use Leibniz formula after differentiating the expression $n$ times (to get some recursive formula):
$$y^{(n)}+2\sum_{k=0}^{n} {n\choose k}{x^{(k)}y^{(n-k)}}=0$$
The sum expression holds only for $k=0,1$ and for $k\ge 2$ it will be $0$ because you differentiate $x$ more then once.
So, it yields, after some calculation: 
$$y^{(n)}+2(xy^{(n-1)}+(n-1)y^{(n-2)})=0$$
You are looking for the $100$-th derivative at $x=0$ so plug it in and you get
$$y^{(n)}(0)=-2(n-1)y^{(n-2)}(0)$$
Now it is over:
$$y^{(100)}(0)=-198y^{(98)}(0)=198\times194\times y^{(96)}(0)=\cdots =\prod_{k=1}^{50}({2+4(k-1)})\times y(0)=\prod_{k=1}^{50}({2+4(k-1)})$$
The product expression you get is actually the expression for arithmetic sequence $2, 6, 10, 14, \ldots , 194, 198$ and it is easy to find that $198$ is $50$-th element of the sequence (that's also why is the expression positive).
