$\left(1+\frac1n\right)^n$ is increasing and bounded without binomial theorem. Title is self-explicative. I have to teach my pupils this sequence to define number $e$, but they don't know yet the binomial theorem. Let's not speak about AM-GM, derivatives, etc.
Is there any way to prove the monoticity and boundness of the sequence with very elementary resources?

We can show monotonicity using Bernoulli's inequality: $(1+r)^k>1+kr$ for $r>-1,k\in\Bbb N,k\geq 2$, which can be proven easily by induction on $k$.
Indeed, assume $n>1$ and take $n=k,r=-\frac{1}{n^2}$. The inequality tells us $(1+\frac{1}{n})^n(1-\frac{1}{n})^n=(1-\frac{1}{n^2})^n>(1-\frac{1}{n})$ which implies $(1+\frac{1}{n})^n>(1-\frac{1}{n})^{1-n}=(1+\frac{1}{n-1})^{n-1}$.
Edit: I believe I got boundedness too. Define an auxilary sequence $(1+\frac{1}{n})^{n+1}$. We will show it's decreasing. We want to prove $(1+\frac{1}{n})^{n+1}<(1+\frac{1}{n-1})^{n}$ or equivalently $(1+\frac{1}{n})^{n}(1+\frac{1}{n})<(1-\frac{1}{n})^{-n}$, $(1+\frac{1}{n})^{n}(1-\frac{1}{n})^{n}<\frac{1}{1+\frac{1}{n}}$, $(1-\frac{1}{n^2})^n<\frac{n}{n+1}$, $(1+\frac{1}{n^2-1})^n>1+\frac{1}{n}$.
We have by Bernoulli that $(1+\frac{1}{n^2-1})^n>1+\frac{n}{n^2-1}>1+\frac{n}{n^2}>1+\frac{1}{n}$ which is what we wanted. (I'm not exactly sure if all transitions here are valid, if anything is wrong or unclear comment)
So thus we have $(1+\frac{1}{n})^n<(1+\frac{1}{n})^{n+1}<(1+\frac{1}{1})^{1+1}=4$. So the sequence is bounded.
