e^x > x^a , a > 1

Hi,

I have this problem:
For all a > 1, fixed, find x_0(a) such that for all x > x_0:
e^x > x^a

Obviously, x_0 exists, and now I think I could prove that we could
take x_0 = e^(2a^2)
But I need some "little" bound. What's the lowest lowest bound of x_0
we take?

Any hints?

Thanks,
Xan.
.