Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$ I need to find an expression for $n$th derivative of $f(x) = e^{x^2}$. Really need help.

Let $y=e^{x^2}$. We want to find $\frac{d^n y}{dx^n}$.
Note that $\ln y = x^2$ and so $\frac{1}{y} \frac{dy}{dx} = 2x\iff \frac{dy}{dx}=2xy$. 
Differentiating again yields:
$\frac{d^2y}{dx^2}=2x\frac{dy}{dx}+2y$,
$\frac{d^3 y}{dx^3} = 2x \frac{d^2 y}{dx^2} + 4\frac{dy}{dx}$,
$\frac{d^4 y}{dx^4} = 2x \frac{d^3 y}{dx^3} + 6\frac{d^2 y}{dx^2}$.
We might guess that $\frac{d^n y}{dx^n} = 2x \frac{d^{n-1} y}{dx^{n-1}}+2(n-1) \frac{d^{n-2} y}{dx^{n-2}}$, which is easily proved by straightforward induction (we leave this as an exercise to the reader. ;) )
Let $u_k=\frac{d^k y}{dx^k}$ for easier notation. We thus have established
$u_n = 2x u_{n-1}+2(n-1)u_{n-2}$.
It is easy to find that the coefficient of $x^n y$ in $u_n$ is $2^n$, but I'm not seeing any nice way of solving the recursion at the moment.
