Proof by induction? There is a problem that I would like to solve. But I am not so sure how. The question is as follows:

Let $r$ be a real number with $0<r<1$. Consider the sequences of rational numbers $a_1,a_2,a_3,\dots$ and $b_1,b_2,b_3,\dots$ given by
  $$a_n= \bigg(1+\frac{r}{2^n}\bigg)^{2^n} , b_n=\bigg(1-\frac{r}{2^n}\bigg)^{2^n}$$
  Prove that $a_{n+1}>a_n$ for all $n\in\mathbb{N}$, and also prove that $b_n\geq 1-r$ for all $n\in\mathbb{N}$. 

A hint is given for the latter part of the problem where Bernoulli's inequality could be used to solve the proof.
At first glance, I think that perhaps this is a proof by induction problem, but then I look at the question and there are $r$ and $n$ which makes me confused.
 A: As $r>0$ is fixed, it plays no role in any induction argument.
Another general comment before coming to the hints. Doesn't those limits remind you somewhat of the limit definition of $e$? 

Hint for a) Study the ratio
$$\frac{a_{n+1}}{a_n}$$

Hint for b) As $r\in (0,1)$, the terms $-r/2^n$ belong to $(-1,1)$. What does Bernoulli's inequality say?
A: By the  AGM inequality (note that $0<1+\frac{r}{2^n}\not=1$ if $0<|r|<1$):
$$GM\left(\underbrace{1,\dots,1}_{\text{$2^n$ times}},\underbrace{1+\frac{r}{2^n},\dots,1+\frac{r}{2^n}}_{\text{$2^n$ times}}\right)< AM\left(\underbrace{1,\dots,1}_{\text{$2^n$ times}},\underbrace{1+\frac{r}{2^n},\dots,1+\frac{r}{2^n}}_{\text{$2^n$ times}}\right).$$
That is
$$c_n(r)^{1/2^{n+1}}=\left(\left(1+\frac{r}{2^n}\right)^{2^{n}}\right)^{1/2^{n+1}}<
\frac{2^{n}+2^n\left(1+\frac{r}{2^n}\right)}{2^{n+1}}=c_{n+1}(r)^{1/2^{n+1}}$$
where $c_n(r)=\left(1+\frac{r}{2^n}\right)^{2^{n}}$,
which implies that 
$$c_n(r)<c_{n+1}(r).$$
Now note that if $0<r<1$, then $a_n(r)=c_n(r)$, and $b_n(r)=c_n(-r)$. Therefore we showed that $a_n<a_{n+1}$ AND $b_n<b_{n+1}$.
Finally, we see that for  $n\geq 1$,
$$b_n>b_0=1-r.$$
A: $$a_{n+1}=\left(1+\frac r{2^{n+1}}\right)^{2^{n+1}}=\left(\left(1+\frac r{2^{n+1}}\right)^2\right)^{2^n}=\left(1+\frac r{2^{n}}+\frac{r^2}{2^{2n}}\right)^{2^n}>a_n.$$
