<s>[INST] 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?
 [/INST] This is the technique that I would use...
Induction Step: We assume $$ \left ( 1 + \frac{1}{k} \right ) ^ k < k. $$
Multiplying both sides by "${\color{red}{something}}$" to get our desired outcome gives,
$$ \left ( 1 + \frac{1}{k} \right ) ^ k {\color{red}{\frac{\left( 1 + \frac{1}{k+1} \right )^{k+1}}{\left ( 1 + \frac{1}{k} \right ) ^ k}}} < k{\color{red}{\left(\frac{k+1}{k}\right)}}. $$
So all you need to show (for this inequality to hold) is 
$$\frac{\left( 1 + \frac{1}{k+1} \right )^{k+1}}{\left ( 1 + \frac{1}{k} \right ) ^ k} < \left(\frac{k+1}{k}\right). $$
Which  is simple enough with some algebra.
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