There are two things that make it difficult to see the laser: 1) the large area illuminated (as BarsMonster pointed out) and 2) the resolving power of the eye.
The eye has an angular resolution of roughly 10^-4 radians (from the diffraction limit), so your eye will "blur" the light from the laser with all the light from the moon's surface within an angle of 10^-4. Using the earth-moon distance of 4e8 m, that means you're competing with the light from a 10^9 m^2 area. Since you say it's a fully lit area of the moon, I'll take an intensity of 1 kW/m^2 for the solar intensity and assume the moon is white (a bad assumption) for a total power of 10^12 W.
The light from the moon's surface is scattered into something like 2-pi of solid angle (as opposed to the laser). So the fraction of light that will be competing with the laser (assuming the laser diameter at the earth is 4000 km) is roughly ((4000 km)/(400000 km))^2= 10^-4.
So if we want the intensity from the laser to equal the intensity of the unresolved scattered light from the moon (which, as a ballpark figure, should make it noticeably green and thus "easily seen"), we need a laser that's 10^8 Watts.
Good luck building that.