Proof of e^x

I think you meant to say this

This has problems I believe:

Set $e^x = x + 1.$

Then, $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = 1 + x$

So, $\frac{x^2}{2!} + \frac{x^3}{3!} + ... = 0$

This only has solution x = 0 (this is where the problem potentially is). Hence, either $e^x \geq x + 1, \forall x$ or $e^x \leq x + 1, \forall x$ because otherwise there would be two points of intersection.

Since $e^x > x + 1$ for $x = -2$, $e^x \geq x + 1 \forall x$.

Alternatively, the fact that $e^x$ is convex and meets x + 1 (which is not convex) at only one point works too.

$x\mapsto x+1$ is convex. What you mean is that it is affine. In fact, if it were not convex, this argument wouldn't work.

I guess I meant strictly convex but yes the fact that it is affine. Why does it need to be convex? If it were concave, wouldn't that also mean that the convex function is above (or equal to) the concave function? I added that argument as an after-thought, I haven't done any of this sort of maths (MVT, concave, convex) for a few years now so I shouldn't really be trying to argue with it at all.

Does anyone know how to prove that for all x>0,

e^x > 1+x

using just alevel maths and further maths knowledge. Ive seen proofs which consider f(x) = e^x-x-1 and show that it is an increasing function etc... But they then go on to use the mean-value theorem which i am not familiar with.

[Btw i am writing this from memory so could anyone clarify if it is either of these or could you have both really]:
e^x> 1+x for x>0
or e^x> or = to 1+x, for x> or equal to 0 (with equality at x=0).

Also, is this true for all x (so now considering negatives; on a quick google search people specify x>0 but e^x gets small very quick for as x gets more negative but is always greater than 0 so i can see it clearly applying for x<-1.What about -1<x<0 then.

I can see in the increasing fucntion proof that f(x) is only increasing for x>0 so i get that restriction but is there another proof which doesnt require that, because i thought it would be true for all x.