<s>[INST] How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$ 
show  that
  $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$

it's well know that 
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand
 [/INST] \begin{align}
(1+\dfrac{1}{16})^{16} &= \sum_{k=0}^{16} {16 \choose k}(\frac{1}{16})^k \\
& = 2 + \frac{15}{32}  + \frac{35}{256} + \sum_{k=4}^{16} {16 \choose k}(\frac{1}{16})^k \\
& \leq 2 +  \frac{15}{32} + \frac{35}{256} +\sum_{k=4}^{16}  \frac{1}{k!}\\
& \leq 2+  \frac{15}{32} + \frac{35}{256} + e - 1 - 1- \frac{1}{2} - \frac{1}{6}\\
& = e - \frac{2}{3} + \frac{155}{256} \\
& \leq 2.719 - 0.666 + 0.606  = 2.659
\end{align}
I used the fact ${n \choose k} \leq \dfrac{n^k}{k!}$ and $e \geq \sum_{k=0}^{16}\dfrac{1}{k!}$. In addition, $e< 2.719, \frac{2}{3} > 0.666, \frac{155}{256} < 0.606$
Added: for a proof which doesn't use the value of $e$, we could use
\begin{align}
\sum_{k=4}^{16}  \frac{1}{k!} \leq \frac{1}{4!}(1 + \frac{1}{5} + \frac{1}{5\times6} +\frac{10}{5\times 6\times 7}) = \frac{269}{7!} < \frac{39}{6!}<  \frac{7}{5!} = \frac{7}{120} < 0.06
\end{align}
Then we have $2 + \frac{155}{256} + \frac{7}{120} < 2 + 0.606 + 0.06 = 2.666$
</s>