<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] As canaaerus points out in the comments, we can in fact prove
$$\left(1+\frac{1}{n}\right)^n<3$$
In fact there's a proof of this that doesn't even use induction! I think it's worth writing down here since it's so simple (and it won't be giving away the answer to the homework problem, since the teacher presumably expects induction). We will binomially expand and use the following facts:
$${{n}\choose{r}}=\frac{n\cdot(n-1)\cdot\ldots\cdot(n-r+1)}{r!}\leq\frac{n\cdot n\cdot\ldots\cdot n}{r!}=\frac{n^r}{r!}$$
and (for $r\geq1$)
$$\frac{1}{r!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot\ldots\cdot r}\leq\frac{1}{1\cdot2\cdot2\cdot2\cdot\ldots\cdot 2}=\frac{1}{2^{r-1}}$$
Here's the proof:
$$\begin{align*}
\left(1+\frac{1}{n}\right)^n&=\sum^{n}_{r=0}{{n}\choose{r}}\frac{1}{n^r}\\
&\leq\sum^{n}_{r=0}\frac{n^r}{r!}\frac{1}{n^r}\\
&=\sum^{n}_{r=0}\frac{1}{r!}\\
&=1+\sum^{n}_{r=1}\frac{1}{r!}\\
&\leq1+\sum^{n}_{r=1}\frac{1}{2^{r-1}}\\
&<1+\sum^{\infty}_{r=1}\frac{1}{2^{r-1}}\\
&=1+2=3
\end{align*}$$
</s>