Two-sided bounds for the logarithm

The inequalities. By the mean value theorem there is $c\in(1,1+x)$ with
$$\ln(1+x)-\ln1=\frac1c\,x.$$
Since $1<c<1+x$, we have $\frac1{1+x}<\frac1c<1$. Multiplying by $x>0$,
$$\frac x{1+x}<\ln(1+x)<x.$$

Deduction. Take $x=\frac1n$. Then $\frac x{1+x}=\frac1{n+1}$ and
$$\frac1{n+1}<\ln\!\left(1+\frac1n\right)<\frac1n.$$
Multiplying by $n$ in the right inequality: $n\ln(1+\frac1n)<1$, so $\left(1+\frac1n\right)^n<e$. Multiplying by $n+1$ in the left inequality: $1<(n+1)\ln(1+\frac1n)$, so $e<\left(1+\frac1n\right)^{n+1}$.

Remark. These bounds squeeze $e$ between two sequences that converge to it: for $n=10$ we get $2.5937<e<2.8531$.