Proving that $\bigl(1+\frac{r}{2^n}\bigr)^{2^n}<\frac{1}{1-r}$ by induction

The question sets a condition where $r\in\mathbb{R}$ and $0<r<1$ with a sequence of rational numbers $a_1,a_2,a_3,\dotsc$ given by $$a_n=\Bigl(1+\frac{r}{2^n}\Bigr)^{2^n}$$ The question then asks to:

Prove that $a_n<\dfrac{1}{1-r}$ for all $n\in\mathbb{N}$.

How I approached this question is to first put the information I have in place

$$\Bigl(1+\frac{r}{2^n}\Bigr)^{2^n}<\frac{1}{1-r}$$

where we can then take a base case of $n=1$.$$\Bigl(1+\frac{r}{2}\Bigr)^2<\frac{1}{1-r}$$

Here I expanded the equation to look like $$1+r+\frac{r^2}{4}<\frac{1}{1-r}$$

But at this point, I really didn't know how I would continue with proving the result. Any help would be appreciated.

$$\left(1+\frac{r}{2^n}\right)^{2^n}=\sum_{i=0}^{2^n}\frac{1}{2^{ni}}\binom{2^n}{i}r^i \le \sum_{i=0}^{2^n} r^i < \sum_{i\ge 0} r^i = \frac{1}{1-r}.$$
Multiplying by the positive number $(1-r)$ the inequality to be proven becomes
$$(1-r)\left(1+r+\frac{r^2}{4} \right) = 1-r^2 + \frac{r^2}{4}-\frac{r^3}{4}<1\,.$$