Step-by-step

Well, I know very little about trigonometry but maybe you can try using order of operations to start it off. Before that, solve the square root of -3. That would give you -1.732. I'm guessing you have to round? That would be -2. Negative two plus one equals -1.  -1 to the power of 5 gives you -1. I think your answer is 1 (when rounding).  

With questions like this it useful to be familia with the trig ratios of the the angles 30, 45 and 60 degrees, (or their radian equivalents, pi/6, pi/4 and pi/3).

Draw a right-angled triangle with sides adjoining the right-angle 1 and sqrt(3), then Pythagoras will tell you that the length of the hypotenuse is 2.

The angles of the triangle will be 30,60 and 90 degrees.

From that you should be able to read off the sine cosine and tangent (or any of the other three ratios), of 30 or 60 degrees.

In particular cos(60 deg) = 1/2, and sin(60 deg) = sqrt(3)/2.

Now to your actual problem.

The sqrt(3) in the problem should, as they say, 'stick out like a sore thumb'.

If there were a 2 underneath it, it would be the sine of 60 degrees (pi/3 radians).

So,

(1 + sqrt(-3))^5 = (1 + i*sqrt(3))^5 = {2(1/2 + i*sqrt(3)/2)}^5 = 32(cos(60) + i*sin(60))^5.

And now you have to use de'Moivres theorem.

= 32(cos(300) + i*sin(300)).