Yes, that looks like a problem.

I don't have a book with me, but I assume this is the standard way:

Let x=a0. Let y be the additive inverse of x (so that x+y=y+x=0). Since 0 is the additive identity, 0 + 0 = 0.

Now x = a0 = a(0+0) = a0 + a0 = x + x. Then adding y to both sides gives:

x + y = (x + x) + y = x + (x + y) = x + 0 = x. Since x + y = 0, have shown that 0 = x. Thus a0 = 0.

Another way:

a = a(1) = a(0+1) = a0 + a(1) = a0 + a. Then add a's additive inverse (-a, but to avoid confusion I'll call it b... thus a+b=b+a=0) to both sides to get:

a + b = ( a0 + a ) + b = a0 + ( a + b ) = a0 + 0 = a0. But then a0 = a + b = 0. Thus a0 = 0.