Mean Value Theorem Proofing For all $ x>0$ By applying the Mean Value Theorem to the function $f(t)=e^t$ defined on the interval $[0, x]$, where $x &gt; 0$, show that for all $x &gt; 0$, $e^x&gt;1+x$.

What I have done...

$f(t) = e^t$
$f$ is continuous on $[0,x]$ and differentiable on $(0,x)$.

For at least one $C$ in domain $(0,x)$ such that
$(f(b)-f(a))/(b-a) = f'(C)$

$(e^0 - e^x)/(-x) = e^c$

$1-e^x = -xe^c$

How do I continue the proof from here?
By the mean value theorem, we have $c\in[0,x]$ such that

$f(x)-f(0)=f'(c)(x-0)$

That is, $e^x-1=e^cx$

Since $e^x$ is increasing and $c&gt;0$ then $e^c&gt;e^0=1$

Thus $e^x-1=e^cx&gt;x$ and thus $e^x&gt;1+x$
Hint: f'(c) is greater than 1, c in (0,x).