I know that $(1+(1/n))^n \to e$ and $(1-(1/n))^n \to 1/e$ as $n \to \infty$.
Both these sequences are increasing. I saw this by plotting; is there an analytical way to see this? Taking the derivative of $(1+(1/x))^x$ or $x \log (1+1/x)$ gets messy and ends up with a quantity $\log((x+1)/x)-\frac{1}{x+1}$ that I don't know what to do with.
I would like lower bounds on $1+(1/n)$ and $1-(1/n)$ using the exponential. If we don't use the above observation that the sequences are increasing, we have $1+(1/n) \ge e^{1/(2n)}$ for all large $n$ and $(1-(1/n))^{-1} \ge e^{1/(2n)}$ for all large $n$, which basically follow from $e>e^{1/2}$.
However, if we use the fact that $(1-(1/n))^n$ is increasing to $e^{-1}$, I think we also have $(1-(1/n))^{-1} \ge e^{1/n}$ for all $n \ge 1$.
I am wondering what are the strongest statements of the above flavor that we can state (regarding lower bounds for $1+(1/n)$ and $1-(1/n)$).
You do not need derivatives or the binomial theorem or lengthy techniques.
By the AM-GM inequality we have:
$$ 1\cdot\underbrace{\left(1+\frac{1}{n}\right)\cdot\ldots\cdot\left(1+\frac{1}{n}\right)}_{n \text{ times}}< \left[\frac{1+n\left(1+\frac{1}{n}\right)}{n+1}\right]^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}\tag{1}$$ and that proves that the sequence given by $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing.
In a similar way we may prove that the sequence given by $b_n=\left(1+\frac{1}{n}\right)^{n+1}=\left(1-\frac{1}{n+1}\right)^{-(n+1)}$ is decreasing. The claim follows.