$\left(1+\frac1n\right)^n$ 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?
 A: 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.
A: Here's a proof that doesn't use the binomial theorem, but it does use integrals and the definition of $e$.  You might be able to extract a more elementary proof from this by just comparing areas under curves and not mentioning integrals, to get your bound.  
Suppose $1\leq t\leq 1+\frac1n$.  Then $$\frac1{1+\frac1n}\leq\frac1t\leq1$$  Therefore
$$\int_1^{1+\frac1n}\frac1{1+\frac1n}dt\leq \int_1^{1+\frac1n}\frac1tdt\leq\int_1^{1+\frac1n}1\cdot dt$$
Therefore
$$\frac1{n+1}\leq \ln\left(1+\frac1n\right) \leq \frac 1n$$
Therefore
$$e^{1/(n+1)} \leq 1+\frac1n \leq e^{1/n}$$
which implies
$$e\leq\left(1+\frac1n\right)^{n+1}$$
which implies
$$\frac{e}{1+\frac1n}\leq \left(1+\frac1n\right)^n$$
which implies
$$(1+\frac1n)^n\leq e$$.
Therefore
$$\frac{e}{1+\frac1n}\leq\left(1+\frac1n\right)^n\leq e$$
squeezing theorem then implies $\left(1+\frac1n\right)^n\rightarrow e$.
A: It is worth to prove AM-GM first, because it gives a rather short proof:
$$a_n=\left(1+\frac{1}{n}\right)^n\cdot 1\leq \left(\frac{1+n\left(1+\frac{1}{n}\right)}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}=a_{n+1}$$
so the sequence is increasing. However:
$$ a_{n+1}-a_n \leq \frac{a_n}{(n+1)^2},$$
hence, by exploiting telescopic products:
$$ a_n \leq a_1 \prod_{j=2}^{n}\left(1+\frac{1}{j^2}\right)\leq a_1 \prod_{j=2}^{n}\left(\frac{1}{1-\frac{1}{j^2}}\right)=\frac{4n}{n+1}<4. $$
