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?

The following inequality will be needed:
$$\frac{1}{n+1}<\frac{1}{n} \Leftrightarrow 1+\frac{1}{n+1}<1+\frac{1}{n}\\
\Leftrightarrow \left(1+\frac{1}{n+1}\right)^n<\left(1+\frac{1}{n}\right)^n.$$
From the induction hypothesis $\left(1+\frac{1}{n}\right)^n<n$ and the algebraic identity $\left(1+ \frac{1}{n+1}\right)^{n+1} = \left(1+ \frac{1}{n+1}\right)^n\left(1+ \frac{1}{n+1}\right)$,
$$\Rightarrow\left(1+\frac{1}{n+1}\right)^n<n\\
\Leftrightarrow \left(1+ \frac{1}{n+1}\right)^n\left(1+ \frac{1}{n+1}\right)<n\left(1+ \frac{1}{n+1}\right)\\
\Leftrightarrow\left(1+ \frac{1}{n+1}\right)^{n+1}<n+\frac{n}{n+1}$$
Can you take it from there?
