<s>[INST] Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$ Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$. This is driving me crazy... I have plotted the graphs to be sure that the inequality is true, and it is, but I can't seem to show it. Here is what I have so far:
$$\left(1 + \frac{x}{n}\right)^{-n} \le \left(1 + \frac{x}{n}\right)^{-x}$$
since $n \ge x$. I want to then use the fact that $1 + \dfrac{x}{n} \le 2$, but that gives me a $\ge$, not a $\le$, so my first step must be wrong... I don't know what else to do, instead.
 [/INST] I have already posted an answer where I considered asymptotic behavior of inequality as $n\to \infty$. 
Below are  more general conclusions.


Given $0\le x \le n$, show that $\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}$.

First, note that 
$$
0\le x \le n \implies 0 \le \frac{x}{n}\le 1
\implies 1\le \left(1 + \dfrac{x}{n}\right) \le 2
$$
Second, note that for any $\alpha \ge 1$ we have
$$
\alpha \ge 1, \quad x \le n \implies 
\alpha^{x} \le \alpha^{n} \implies 
\alpha^{-x}  = \frac{1}{\alpha^{x}}\ge \frac{1}{\alpha^{n}} = \alpha^{-n}
$$
Third, for $1\le \alpha \le \beta$ and for $n \ge 0$ we have
$$
n\ge 0 \implies  \alpha \le \beta \iff\alpha^{n} \le \beta^{n} \iff 
\alpha^{-n} \ge \beta^{-n} 
$$
Fourth, rewrite $2^{-x}$ as $\left(2^{-\frac{x}{n}}\right)^{n}$, get
$$
2^{-x} = \left(2^{-\frac{x}{n}}\right)^{n}
\implies 
\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x} \iff 
\left(1 + \dfrac{x}{n}\right)^{-n} \le \Big(2^{\frac{x}{n}}\Big)^{-n}
$$
Finally, we only have to show that 
$ \left(1 + \dfrac{x}{n}\right) \ge 2^{\frac{x}{n}}$  whenever $0 \le x \le n$.
Let $y:= \frac{x}{n}\,\ln 2$, then
$$
\begin{aligned}
2^{\frac{x}{n}} = e^{\frac{x}{n}\cdot \ln 2}  = e^{y} 
& = \sum_{k=0}^{\infty}\frac{y^{k}}{k!}
= \sum_{k=0}^{\infty}\frac{\ln^{k}(2)}{n^{k}}\frac{x^{k}}{k!}
= 1 +  \frac{\ln2 }{n}x + \frac{\ln^{2}(2)}{n^{2}}\frac{x^{2}}{2} 
+ \frac{\ln^{3}(2)}{n^{3}}\frac{x^{3}}{6} + \cdots = \\
& = 1 +  \frac{\ln2 }{n}x + \mathcal{O}\left(x^2\right) 
\\
\left(1 + \dfrac{x}{n}\right)  = 1 + \dfrac{1}{n}x
& = 1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right).
\end{aligned}
$$
Since $\ln 2< 1$, we conclude that $ 2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right) $, and thus establish inequality:
$$
\begin{aligned}
\ln 2< 1 &\implies
1 + \dfrac{\ln 2}{n}x + \mathcal{O}\left(x^2\right) \le  1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right)\implies 
\\ & \implies 
2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right)   , \quad 0\le x \le n \implies 
\\ & \implies 
\boxed{\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}}
\end{aligned}
$$
Q.E.D.
</s>