My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.
If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.
$Proof.$ We use mathematical induction.
Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.
Inductive Step. Let $n>1$. Assume our statement is true for $n$. Now observe that
$\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$
My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?
Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
Note that we have to show that $$\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)\left(1-\frac{1}{2^{n+1}}\right)\color{red}{\ge}\frac 14+\frac{1}{2^{n+2}}$$
One has $$\begin{align}\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)\left(1-\frac{1}{2^{n+1}}\right)&= \frac 14-\frac{1}{2^{n+3}}+\frac{1}{2^{n+1}}-\frac{1}{2^{2n+2}}\\&= \frac 14+\frac{-2^{n-1}+2^{n+1}-1}{2^{2n+2}}\\&=\frac 14+\frac{3\cdot 2^{n-1}-1}{2^{2n+2}}\\&= \frac 14+\frac{2^n+(2^{n-1}-1)}{2^{2n+2}}\\&\color{red}{\ge}\frac 14+\frac{2^n+0}{2^{2n+2}}\\&=\frac 14+\frac{1}{2^{n+2}}\end{align}$$