Find the Largest integer that is smaller than this expression Largest integer that is smaller than
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2^0}+1)(2^{2^2}-2^{2^1}+1)...(2^{2^{2020}}-2^{2^{2019}}+1)}$
is...
My Progress
$\frac{1}{2^{2^{k+1}}-2^{2^k}+1}=\frac{2^{2^k}+1}{2^{3.2^k}+1}$
Reorder so
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2^0}+1)(2^{2^2}-2^{2^1}+1)...(2^{2^{2020}}-2^{2^{2019}}+1)}$=$\prod_{i=0}^{2018}\frac{2^{2^{k+1}}}{2^{3.2^{k}}+1}.\frac{3.2^{2^{2021}}}{2^{3.2^{2019}}}< \frac{3.2^{2^{2021}}}{2^{3.2^{2019}}}$
I'm stuck here :(
If someone was able to finish this, that would be great, I've been looking for the answer for more than 1 hour:(

Your first step is, good.
Then use:
$$(x-1)\prod_{k=0}^n \left(x^{2^k}+1\right)=x^{2^{n+1}}-1 $$
Prove by induction.
Use it with $x=2,x=2^3.$
Then $$\begin{align}\frac1{\prod_{k=0}^{2019}\left(2^{2^{k+1}}-2^{2^k}+1\right)}&=\prod_{k=0}^{2019}\frac{2^{2^k}+1}{2^{3\cdot 2^k}+1} \\
&=(8-1)\frac{2^{2^{2020}}-1}{2^{3\cdot 2^{2020}}-1}
\end{align}$$
So $$\begin{align}\frac{2^{2^{2021}}}{\prod_{k=0}^{2019}\left(2^{2^{k+1}}-2^{2^k}+1\right)}&=7\cdot\frac{2^{3\cdot 2^{2020}}-2^{2\cdot 2^{2020}}}{2^{3\cdot2^{2020}}-1}\\
&=7\cdot\left(1-\frac{4^{2^{2020}}-1}{8^{ 2^{2020}}-1}\right)\end{align}$$
And that fraction is much smaller than $\frac17,$ so the greatest integer is $6.$

More generally, $$\frac{x^{2^n}}{\prod_{k=0}^{n-2}\left(x^{2^{k+1}}-x^{2^k}+1\right)}=(x^2+x+1)\left(1-\frac{x^{2^{n}}-1}{x^{3\cdot 2^{n-1}}-1}\right)$$
Also when $x>1,$ $$\frac{x^{2^n}-1}{x^{3\cdot 2^{n-1}}-1}=\frac{x^{2^{n-1}}+1}{x^{2^n}+x^{2^{n-1}}+1}<\frac{1}{x^{2^{n-1}}}$$
In particular, for $x>1$ an integer and $n>2,$ the greatest integer would always be $x^2+x.$
When $n=2$ we have:
$$\frac{x^4}{x^2-x+1}=x^2+x-\frac{x}{x^2-x+1}$$
So the greatest integer is $x^2+x-1$ in that case.
