Proving an inequality which the solution says "is obvious" but I cant see how

Question

I was going through my Lebesgue Measure and Integration course, and I came across this inequality. $$ \left(1+\frac{x}{n}\right)^{-n}\leq e^{-x} $$
I tried expanding both sides, and got
$$\text{LHS}= 1-x+\frac{n+1}{n}\frac{x^2}{2!}-\frac{(n+1)(n+2)}{n^2}\frac{x^3}{3!}\cdots $$
and $$ \text{RHS}=1-x+\frac{x^2}{2!}\cdots $$ Now, I noticed that all the terms in the LHS are either equal or bigger than the corresponding terms in RHS but I don't know how to conclude the inequality from this. Am I missing something extremely simple and obvious here?
Any help will be appreciated.

Answer 1

For your inequality to be true, we need $n<0$ and $x \ge 0$. Note that from this link we know that $$e^{\frac{x}{n}} \ge \;1+\frac{x}{n}$$ Now, as $t^{-n}$ is an increasing function if $n<0$, $t>0$, we exponentiate each side by $-n$ to get $$e^{-x} \ge \bigg{(}1+\frac{x}{n}\bigg{)}^{-n}$$ as desired. Of course, if $n \ge 0$, the inequality signs reverse, for a similar reason. So your inequality is, for example, not true when $n=1$ as stated previously.

Answer 2

We can take a logarithm from both sides. Then, we have $-n \log(1 + \frac{x}{n}) \leq -x \Rightarrow n \log(1 + \frac{x}{n}) \geq x \Rightarrow \log(1 + \frac{x}{n}) \geq \frac{x}{n}$

let $\frac x n = y$. Then, we have

$\log(1+y) \geq y$ and this is not true for any $y$