The remainder when $\left|\begin{array}{ccc}{2014}^{2014}& {2015}^{2015}& {2016}^{2016}\\ {2017}^{2017}& {2018}^{2018}& {2019}^{2019}\\ {2020}^{2020}& {2021}^{2021}& {2022}^{2022}\end{array}\right|$ is divided by 5 is equal to 

$\left|\begin{array}{ccc}{(2015-1)}^{2014}& {2015}^{2015}& {(2015+1)}^{2016}\\ {(2015+2)}^{2017}& {(2020-2)}^{2018}& {(2020-1)}^{2019}\\ {2020}^{2012}& {(2020+1)}^{2021}& {(2020+2)}^{2022}\end{array}\right|$

Since the expression of the form ${(5n-a)}^k$ Leaves the remainder ${(-a)}^k$ when it divided by 5.
Thus, remainder  
$=\left|\begin{array}{ccc}1& 0& 1\\ 2& 4& -1\\ 0& 1& 4\end{array}\right|=19=5.3+4$ remainder  = 4