Finitely Nested Nests

Start with the formula for the sum of a finite geometric series: $\sum_{j=0}^k n^j = \frac{n^{k+1}-1}{n-1}$

Manipulating this, $\begin{aligned} (n-1)\sum_{j=0}^k n^j &= n^{k+1}-1 \\ 1+(n-1)\sum_{j=0}^k n^j &= n^{k+1} \\ \sqrt[k+1]{1+(n-1)\sum_{j=0}^k n^j } &= n \end{aligned}$

In the expression in the question, the final root is $\sqrt[2021]{1+2021\sum_{j=0}^{2020} 2022^j }$

So setting $k=2020$ , $n=2022$ and using the formula above, this works out to be $2022$ . Now $N=\sqrt[2021]{1+\sum_{j=0}^{2020} 4021^j \sqrt[2021]{1+\sum_{j=0}^{2020} 4020^j \sqrt[2021]{1+\sum_{j=0}^{2020} 4019^j \sqrt[2021]{1+\cdots \sqrt[2021]{1+2022\sum_{j=0}^{2020} 2023^j } }}}}$

This process continues, and we find $N=\sqrt[2021]{1+4020\sum_{j=0}^{2020} 4021^j }=\boxed{4021}$