Method 1:
Using the power series for $\log(1+x)$, we get $ \begin{align} \log\left(1+\frac1n\right) &=-\log\left(1-\frac1{n+1}\right)\\ &=\frac1{n+1}+\frac1{2(n+1)^2}+\frac1{3(n+1)^3}+\dots\tag{1} \end{align} $ Multiply $(1)$ by $n+1$: $ (n+1)\log\left(1+\frac1n\right)=1+\frac1{2(n+1)}+\frac1{3(n+1)^2}+\dots\tag{2} $ $(2)$ is obviously decreasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n+1}$ is decreasing in $n$.
Multiply $(1)$ by $n$: $ \begin{align} n\log\left(1+\frac1n\right) &=((n+1)-1)\log\left(1+\frac1n\right)\\ &=1-\frac1{1\cdot2(n+1)}-\frac1{2\cdot3(n+1)^2}-\dots\tag{3} \end{align} $ $(3)$ is obviously increasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n}$ is increasing in $n$.
Therefore, since $\displaystyle e=\lim_{n\to\infty}\left(1+\frac1n\right)^{\large n}$, we have $ \left(1+\frac1n\right)^{\large n}< e<\left(1+\frac1n\right)^{\large n+1}\tag{4} $
Let $n=6$, then $ \frac52<\left(\frac76\right)^{\large 6}< e<\left(\frac76\right)^{\large 7}<3\tag{5} $ Method 2:
Since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}\ $, we have $ \begin{align} \frac52=1+1+\frac12< e &=1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot3}+\frac1{1\cdot2\cdot3\cdot4}+\dots\\ &<1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot2}+\frac1{1\cdot2\cdot2\cdot2}+\dots\\ &=3\tag{6} \end{align} $