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

$$ (1+\dfrac{1}{16})^{16}
 = \sum \frac{1}{16^k}\binom{16}{k} 
=  \sum \frac{1}{ 16^k}\frac{16!}{(16-k)!} \frac{1}{k!} < 1 + 1 + \frac{1}{2} + \frac{1}{6} + \dots <\dfrac{8}{3} + \frac{2}{4!}$$
How do we know the remaining terms are small enough?  Let's try an inequality.  In our case, $n=4$.
$$ \sum_{k=n}^\infty \frac{1}{k!} 
 = \frac{1}{n!}\sum_{k=0}^\infty \frac{1}{n^k} = \frac{1}{n!}\frac{1}{1-\frac{1}{n}}< \frac{2}{n!}$$
If we are more careful we can actually prevent the overshooting that @StevenStadnicki points out.
$$1 + 1 + \frac{1}{2}\frac{15}{16} +   \frac{1}{6}\frac{15\times 14}{16\times 16 } + \frac{1}{12}= 2.688 > \frac{8}{3} = 2.66\overline{6}$$
If you extend out to the 4th term the result is 2.654 which is bigger than 2.6379 which is the exact answer up to 4 digits.
