Is $1-\alpha x\le (1-x)^{\alpha}\le 1-\alpha x+\frac{\alpha(\alpha-1)}{2}x^2,$ an inequality about generalized binomial coeffients true? Is the following inequality true?
$$1-\alpha x\le (1-x)^{\alpha}\le 1-\alpha x+\frac{\alpha(\alpha-1)}{2}x^2$$ for real numbers $x,\alpha.$ We may assume $0\le x\le 1$ and put some requirements about $\alpha.$ And $(1-x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}x^k,$ where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$
I think I might meet with this inequality in Stanley's book Enumerative Combinatorics 1 http://www-math.mit.edu/~rstan/ec/ec1.pdf or somewhere else, saying that $(1-x)^\alpha$ greater the sum of its expansion terms up to a negative sign, and less than sum up to a positive sign. But I cannot remember it now and I am not sure. And it seems natural if we view it as a Taylor expansion. Any proof, comments or reference are very welcome!

For the upper bound, use
\begin{align}
(1-x)^r\leq e^{-rx}= 1-rx+ \frac{r^2x^2}{2!}-\frac{r^3x^3}{3!}+\ldots
\end{align}
for all $x<1$ and $r>0$. 
Next, consider the following fact: for $x<0$, we have
\begin{align}
e^x \leq T_{2n}(x)
\end{align}
for all $n$ where
\begin{align}
T_n(x) = 1+x+\frac{x}{2!}+\ldots +\frac{x^n}{n!}.
\end{align}
Combining both result yields the desired inequality for all $r>0$ and $0<x<1$. 
