Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$ First we need to prove the basis. If we let $n=3$, then $(1+ \frac{1}{3})^3 < 3$ 
$(\frac{3}{3}+ \frac{1}{3})^3 < 3$ 
$(\frac{4}{3})^3 < 3$ 
$(\frac{64}{27}) < 3$ 
The inequality statement is true
For $P(n), (1+ \frac{1}{n})^n < n$
We assume that $(1+ \frac{1}{n})^n < n$ is true for $P(n+1)$
$(1+ \frac{1}{n+1})^{n+1} < n+1$ 
$(1+ \frac{1}{n+1})^{n})(1+ \frac{1}{n+1})^{1}) < n+1$ 
And then I'm stuck afterwards. I know that there are a variety of problems that use induction and they have different methods, but I only know the ones that are similar to $1+2+3+...+n = n+2$  or $7^n-8^n$ is divisible by $8$. Is there any technique to tackle this type of problem?

This is the same as
$n
> (\frac{n+1}{n})^n
$
or
$\frac{(n+1)^n}{n^{n+1}}
< 1
$.
This is true for
$n=3$,
since
$4^3 = 64$
and
$3^4 = 81$.
I will show that
$\frac{(n+1)^n}{n^{n+1}}
$
is decreasing;
since this is less than one
for $n=3$,
this will show that
it is less than one
for all $n \ge 3$.
Showing this is decreasing
is the same as showing that
$\frac{(n+1)^n}{n^{n+1}}
> \frac{(n+2)^{n+1}}{(n+1)^{n+2}}
$.
Cross-multiplying,
this is the same as
$(n+1)^{2n+2}
>(n(n+2))^{n+1}
$
or
$(n^2+2n+1)^{n+1}
> (n^2+2n)^{n+1}
$
which is obviously true.
