How to prove $e^t-1 > (e-1) t^2$. I want to prove 

$$e^t-1 \ge (e-1) t^2, \qquad t\in[0,1].$$

For $t\in[0,1]$, we have $t^n \le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$
 A: Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $t\in[0,1]$,
$$f''(t)=e^t-2(e-1)\leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $t\in[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+t\cdot 1)\geq f(0)+t\cdot f(1)=0.$$
A: The inequality is trivially an equality for $t=0$ or $t=1$.  Define $$f(t):=\big(\exp(t)-1\big)-(\text{e}-1)\,t^2=\sum_{k=1}^\infty\,\frac{t^k}{k!}-\sum_{k=1}^\infty\,\frac{t^2}{k!}=(t-t^2)-\sum_{k=3}^\infty\,\frac{t^2(1-t^{k-2})}{k!}$$
for $t\in(0,1)$.
That is,
$$g(t):=\frac{f(t)}{t(1-t)}=1-\sum_{k=3}^\infty\,\frac{t\,\left(1+t+t^2+\ldots+t^{k-3}\right)}{k!}> 1-\sum_{k=3}^\infty\,\frac{(k-2)}{k!}$$
for every $t\in(0,1)$.
Now,
$$\sum_{k=3}^\infty\,\frac{(k-2)}{k!}=\sum_{k=3}^\infty\,\frac{1}{(k-1)!}-2\,\sum_{k=3}^\infty\,\frac{1}{k!}=\left(\text{e}-2\right)-2\left(\text{e}-\frac{5}{2}\right)=3-\text{e}\,.$$
That is,
$$g(t)> 1-(3-\text{e})=\text{e}-2>0\,.$$
The claim immediately follows.  In fact, this proof gives a stronger inequality:
$$\exp(t)\geq 1+ (\text{e}-1)\,t^2+(\text{e}-2)\,t(1-t)=1+(\text{e}-2)\,t+t^2\text{ for every }t\geq 0\,.$$
A: Proof
Denote $$f(x)=x^t,~~~~~t \in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(\xi)(e-1),~~~\xi \in (1,e).$$
Namely, $$e^t-1=t\xi^{t-1}(e-1).$$
Thus, $$\xi^{t-1} \geq e^{t-1} \geq t\geq 0,$$which is desired.
A: Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=\frac{1\pm\sqrt{1-4z}}2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^{\frac{1\pm\sqrt{1-4z}}2}-1-(e-1)\left(\frac{1\pm\sqrt{1-4z}}2\right)^2$$
seems to have a Taylor development with positive-only coefficients.

But this needs to be proven.
