Question on the proof of $e^x>1+x$ for $x>0$ 
Show that $e^x>1+x$ for $x>0$
Proof: Set $f(x)=e^x-(1+x)$. Show that $f(x)$ is always positive. We know that $f(0)=e^0-(1+0)=0$ and $f'(x)=e^x-1$. When $x$ is positive, $f'(x)$ is positive because $e^x>1$
We know that if $f'(x)>0$ on an interval, then $f(x)$ is increasing on that interval, so we can conclude that $f(x)>f(0)$ for $x>0$ and thus:
$$e^x-(1+x)>0 \iff e^x>1+x$$

Question: How do we know that $f(x)$ is positive just by $e^x>1$?
Additional question, how do we prove $f'(x)$ is positive/negative over an interval $[a,b]$?

I believe you are over-complicating it. $e^x$ is positive and log-convex, hence convex. In particular its graph lies above any tangent line, like, for instance, the tangent line at $(0,1)$, whose equation is $y=x+1$.
