<s>[INST] Proving $(1 + \frac{1}{n})^n < n$ for natural numbers with $n \geq 3$. Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.
Attempt at proof: Basic step. This can be verified easily. 
Induction step. Suppose the assertion holds for $n >3$, then we now prove it for $n+1$. We want to prove that \begin{align*} \big( 1+ \frac{1}{n+1})^{n+1} < n+1. \end{align*} So we have \begin{align*} \big( 1+ \frac{1}{n+1}\big)^{n+1} &= \big( 1+ \frac{1}{n+1} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & < \big (1 + \frac{1}{n} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big)  \\ & = n \cdot \big( 1 + \frac{1}{n+1} \big) \qquad \text{(Induction hypothesis)} \\ &= n \cdot \big( \frac{n+1+1}{n+1} \big) \\ &= \frac{n^2 + 2n}{n+1} \\ &= \frac{(n+1)^2 -1}{(n+1)}
\end{align*}
And now I'm stuck. I don't know how to get $n+1$ on the RHS. Please help!
 [/INST] For $n=3$ we have
$(1 + 1/n)^{n} < n$.
If $n \geq 1$ such that
$(1 + 1/n)^{n} < n$,
then
$$
(1 + \frac{1}{n+1})^{n+1} < (1 + \frac{1}{n})^{n+1} < n(1 + \frac{1}{n}) = n+1,
$$
qed.
</s>