Let $\displaystyle f_n(x) = e^x - \left( 1 + x + ... + \frac{x^n}{n!} \right)$.
You need to prove that $\displaystyle f_n(x) > 0$.
You can prove this by showing $\displaystyle f ' (x) > 0\text{ for }x>0$ and $\displaystyle f(0) \geq 0$.
Prove this by induction.
What you need to realize is that $\displaystyle f_{n+1}(x) ' = f_n(x)$.
First $\displaystyle f_1(0) = e^0 - (1 + 0) = 0$. Second, $\displaystyle f_1'(x) = e^x - 1 > 0$ for $\displaystyle x>0$. Thus, the function $\displaystyle f_1$ is increasing for $\displaystyle x>0$. But at $\displaystyle x=0$ we have $\displaystyle f_1(0) = 0$ which means if $\displaystyle x>0$ then $\displaystyle f_1(x) > f_1(0) = 0$. Therefore, $\displaystyle e^x - (1+x) > 0 \text{ for }x>0$.
How many boards are you posting this on? I feel like I am chasing you around the internet! I explained this to you on "Physics Forums" but since you might confuse others:
You are NOT asked to prove that "this function is larger than its Taylor series". You are asked to prove that it is larger than its Taylor polynomial of degree n, for any n.
the remainder formula is
0<c<x
$\displaystyle
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$
so i guess that for x>0 its:
$\displaystyle
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$
for x<0
$\displaystyle
R_n(x)=-\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}
$
should i prove it by induction??
if so then if n=1 then
$\displaystyle
e^x=1+x+ \frac{f^{1+1}(c)}{(1+1)!}x^{1+1}
$
but i dont know whats the value of c??
and still it depends on the values of x
??