$(2021{)}^{2020}=(1+2020{)}^{2020}$

${}^{2020}\mathrm{C}_0+{}^{2020}\mathrm{C}_1·2020+{}^{2020}\mathrm{C}_2·{2020}^2+\dots +{}^{2020}\mathrm{C}_{2020}·{2020}^{2020}$

$1+{\left(2020\right)}^2+{}^{2020}\mathrm{C}_2·{2020}^2+\dots +{}^{2020}\mathrm{C}_{2020}·{2020}^{2020}$

$1+{\left(2020\right)}^2\left(1+{}^{2020}\mathrm{C}_2+\dots +{\left(2020\right)}^{2018}\right)$

$1+(2020{)}^2·\lambda$

Hence $(2021{)}^{2020}=\lambda (2020{)}^2+1$

Hence remainder $=1$