Q:

Proof of Reduction Formula for $\int_{0}^{\infty} x^n e^{-x^2} dx$

I have attempted both substitution and integration by parts upon the following problem (and both in conjunction) to no avail. The reduction formula given remains quite elusive. 
Let $$I_n=\int_{0}^{\infty} x^ne^{-x^2}dx.$$
Hence, show that $$I_n = \frac{n-1}{2}I_{n-2}, \ n \geq 2.$$
I have the following:
$$I_n = \int_{0}^{\infty} x^ne^{-x^2} dx \\ = \left[-\frac{1}{2}e^{-x^2}x^{n-1}\right]_0^\infty - \int_{0}^{\infty} -\frac{1}{2}e^{-x^2}(n-1)x^{n-2}dx \\ =  \left[-\frac{1}{2}e^{-x^2}x^{n-1}\right]_0^\infty + \frac{1}{2(n-1)} \int_{0}^{\infty} e^{-x^2}x^{n-2}dx \\ = \left[-\frac{1}{2}e^{-x^2}x^{n-1}\right]_0^\infty + \frac{1}{2(n-1)}I_{n-2} \\ = \frac{1}{2(n-1)}I_{n-2} \ \ \mathrm{due \ to \ fact \ that:} \ \  \lim_{x \to \infty}x^{n-1}e^{-x^2}=0$$
Which is not quite the solution. Where have I erred? 

A:

HINT:
When integrating by parts, let $u=x^{n-1}$ and $v=-\frac12 e^{-x^2}$.  
Also, use the fact that for all $n$
$$\lim_{x\to \infty} x^{n-1}e^{-x^2}=0$$

