The following is from Tom Apostol's Calculus I, on page 250, exercise 42.:
If $\mathit{n}$ is a positive integer and if $\mathit{x} > 0$, show that
$$ \left( 1 + \frac{x}{n} \right)^n < e^x \text, \qquad \text{and that} \qquad e^x < \left(1 - \frac{x}{n}\right)^{-n} \quad \text{if} \quad x < n. $$
There is no solution provided in the book, I would like to ask someone to verify if mine is correct:
Integrating the inequality $ 1 > \frac{1}{1 + \frac{x}{n}} $ we get that
$$ \int_0^x{1dt} > \int_0^x{\frac{1}{1 + \frac{t}{n}}dt} \qquad \text{yielding} \qquad x > n\log\left(1 + \frac{x}{n}\right). $$
Since $ e $ is strictly increasing it follows that $$ e^x > \left(1 + \frac{x}{n}\right)^{n}. $$
Similarly, integrating $ 1 < \frac{1}{1 - \frac{x}{n}} $ we get that $$ \int_0^x{1dt} < \int_0^x{\frac{1}{1 - \frac{t}{n}}dt} \qquad \text{yielding} \qquad x < -n\log\left(1 - \frac{x}{n}\right) . $$ It follows that $$ e^x <\left(1 - \frac{x}{n}\right)^{-n} . $$
Your proofs look fine.
Another approach is to use the fact that $e^x\ge1+x$ for all $x\in\mathbb{R}$, with strict inequality for $x\ne0$. This follows from the strict convexity of $e^x-1-x$ and its minimum of $0$ at $x=0$.
Then we have for $x\gt0$, $$ e^{x/n}\gt1+\frac xn\implies e^x\gt\left(1+\frac xn\right)^n\tag{1} $$ and $$ e^{-x/n}\gt1-\frac xn\implies e^x\lt\left(1-\frac xn\right)^{-n}\tag{2} $$ Inequality $(2)$ does rely on the fact that $x\lt n$ so that the quantity being raised to the negative power is positive.