Let f(x)=e^x-x-1

This function is differenciable apply the increasing/decreasing theorem,

f'(x)=e^x-1

Now if it true that e^x-1>0 on x>0.

Yes, why?

Because, it is equivalent to saying,

e^x>1 and e^0=1 thus if x>0 then e^x > 1 because e^x is an increasing function. Q.E.D.