Inequality: powers of small numbers If $\epsilon \approx 0 $ then which one is greater $1-\epsilon^{k}$ or $(1-\epsilon)^{k}$ where $k \in \mathbb{Z}_{> 0}$ is a positive integer.
 A: Assume $\epsilon > 0, k \in \mathbb{N}\implies f(\epsilon) = 1-\epsilon^k - (1-\epsilon)^k\implies f'(\epsilon) = -k\epsilon^{k-1}+k(1-\epsilon)^{k-1}> 0$ since $\epsilon \approx 0\implies \epsilon < \frac{1}{2}\implies 1-\epsilon > \epsilon\implies f'(\epsilon) > 0\implies f(\epsilon) > f(0)=0\implies 1-\epsilon^k > (1-\epsilon)^k$.
A: If $k=1$ the two expression are clearly equal.
If $k\geq 2$, using the binomial theorem, we find that
$$(1+\epsilon)^k=1-k\epsilon+\mathcal{o}(\epsilon).$$
Where we have used that if $k\ge2$, $\epsilon^k=\mathcal{o}(\epsilon)$. By the same reason the other expression is of the form
$$1+\mathcal{o}(\epsilon).$$
Thus, because of this asymptotic behavior, for small $\epsilon$ the first expression will be smaller.
Rmk: Check out @DeepSee's answer, it is quite nice how he manages to quantify explicitly the inequality.
A: In the same spirit as DeepSea's answer, consider $$f(x) = 1-x^k - (1-x)^k$$ $$f'(x)=k (1-x)^{k-1}-k x^{k-1}$$ $$f''(x)=-(k-1) k \left(x^{k-2}+(1-x)^{k-2}\right)$$ The first derivative cancels at $x=\frac 12$ for which $$f\left(\frac 12\right)=1-2^{1-k}\qquad , \qquad f''\left(\frac 12\right)=-2^{3-k} (k-1) k$$ The second derivative shows that this is a maximum the value of which being always positive if $k>1$.
So, for any $0 <x<1$ and $k >1$, we always have $$1-x^k > (1-x)^k$$
A: If we take that $a \approx 0$ and $a>0$ and $k \geq 2$ and suppose that $(1-a)^k \geq 1-a^k$ then we arrive at $(1-a)^k \geq (1-a)(1+a+...+a^{k-1)}$ from which it follows that $(1-a)^{k-1} \geq 1+a+...+a^{k-1}$ which is clearly false because the left hand side is less than one and the right hand side is greater than one so we arrived at the contradiction so we have $(1-a)^k<1-a^k$.
