Dick said:
I'm not sure that helps. The OP could have graphed e^x-1-x to begin with. How do you even define 'e' without calculus?
I have absolutely no idea

. I've been told I'm not allowed to use anything with haven't learned in the course to this point (although the Professor busted out the Triangle Inequality recently, which was definitely not covered in any lecture), and I would be willing to wager that the IMV is out as well. Is there some way I can state it simply and internally?
In our course so far, we've spent most of our time on the least upper bound property and basic properties of sequences. I'm really stumped :(.
Anyways, at this moment I've given up on an awesome solution, so I plan on handing something vaguely passable in, and here's where I am at for now:
e^x\geq1+x for all x\geq0
equivalent to e^x-1-x\geq0 for all x\geq0
Proposition i: There is no x\geq0 s.t. e^x-1-x<0
Proposition ii: Proposition i is false:
Suppose Proposition ii is true.
Then there exists x greater than or equal to 0 such that e^x-1-x < 0
<=> e^x-x < 1
<=> ln(e^x-x) < ln(1)
<=> ln(e^x-x) < 0
ln(y)<0 does not exist. Thus proposition ii is false, and thus proposition i is true.
What do you think? I'm probably just going to hand this in, so if it is junk, I'd love to know about it, if only for future reference.